RD Sharma Class 12 Exercise 30.7 Probability Solutions Maths

RD Sharma Class 12 Exercise 30.7 Probability Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 25, 2022 04:02 PM IST

RD Sharma class 12th exercise 30.7 is undoubtedly one of the best NCERT solutions, which has been pretty much decided by high school students. These solutions are not your regular run-of-the-mill books that don't have a lot of information. The book is rich with new and modern calculations which are known to only a few experts. RD Sharma solutions These answers will hugely benefit high school students who struggle to find good study material for exam preparations. RD Sharma class 12 chapter 30 exercise 30.7 will end this tiring search and won't even require them to invest in any redundant study materials.

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RD Sharma Class 12 Solutions Chapter30 Probability - Other Exercise
Probability Excercise: 30.7
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RD Sharma Class 12 Solutions Chapter30 Probability - Other Exercise

Probability Excercise: 30.7

Probability exercise 30.7 question 2

Answer:
$\frac{25}{52}$
Hint:
Use baye’s theorem.
Given:
A bag A contained 2 white and 3 red ball and a bag B contain 5 red and 4 white ball one ball drawn at random from one of bag is found red. What is probability that it was drawn from bag B?.
Solution:
The event selecting red ball denoted by R.
The event selected by bag A denoted by A.
The event selected by bag B denoted by B.
Bag A = 2 white and 3red ball.
Bag B = 4 white and 5 red ball.
$\begin{aligned} &P(A)=P(B)=\frac{1}{2}\\ &P\left ( \frac{R}{A} \right )=\frac{3}{5}\\ &P\left ( \frac{R}{B} \right )=\frac{5}{9}\\ \end{aligned}$
Using Baye’s theorem
$\begin{aligned} &P\left ( \frac{B}{R} \right )=\frac{P\left ( \frac{R}{A} \right ).P(B)}{P(R)} \\ &=\frac{P\left ( \frac{R}{A} \right ).P(B)}{P\left ( \frac{R}{A} \right ).P(A)+P\left ( \frac{R}{B} \right ).P(B)} \\ &=\frac{\frac{5}{9}.\frac{1}{2}}{\frac{3}{5}\times \frac{1}{2}+\frac{5}{9}\times \frac{1}{2}}\\ &=\frac{25}{52} \end{aligned}$

Probability exercise 30.7 question 3

Answer:
$\frac{2}{9}$
Hint:
Use baye’s theorem.
Given:
Three win contain 2 white and 3 black balls; 3 white, 2 black and 4 white and 1 black balls. One ball drawn from a win chosen at random and it was found to white. Find the probability of win drawn.
Solution:
Let E₁,E₂,E₃ denote the event of selecting win I, win II, win III.
Let A be event that ball drawn white.
$\begin{aligned} &P(E_1)=\frac{1}{3},\\ &P(E_2)=\frac{1}{3},\\ &P(E_3)=\frac{1}{3}\\ &P\left ( \frac{A}{E_1} \right )=\frac{2}{5}\\ &P\left ( \frac{A}{E_2} \right )=\frac{3}{5}\\ &P\left ( \frac{A}{E_3} \right )=\frac{4}{5}\\ \end{aligned}$
Using Baye’s theorem we get
Required probability
$\begin{aligned} &P\left (\frac{E_1}{A} \right )=\frac{P(E_1).P\left ( \frac{A}{E_1} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )+P(E_3)\times P\left ( \frac{A}{E_3} \right )}\\ &=\frac{{\frac{1}{3}\times \frac{2}{5}}}{\frac{1}{3}\times \frac{2}{5}+\frac{1}{3}\times \frac{3}{5}+\frac{1}{3}\times \frac{4}{5}}\\ &=\frac{\frac{2}{5}}{\frac{2+3+4}{5}}\\ &=\frac{2}{9} \end{aligned}$

Probability exercise 30.7 question 5

Answer:
$\frac{8}{11}$
Hint:
Use baye’s theorem.
Given:
Suppose a girl throw a die. If she gets 1, 2 she toss a coin three times and note the number of tails. If she get 3, 4, 5, 6 she toss a coin once and note whether a head or tail is obtained. If she obtained exactly one tail, what is probability she threw 3, 4, 5, 6 with die.
Solution:
Let E₁ be event that girl get 1 or 2.
Thus
$P(E_1)=\frac{2}{6}=\frac{1}{3},P(E_2)=\frac{4}{6}=\frac{2}{3},P(E_3)=\frac{3}{8}$
Let E₂ be event that obtain tail. If she tossed a coin 3 time and exactly 1 tail show up then total favorable.
{ (THH), (HTH), (HHT) }=3
$P\left ( \frac{A}{E_1} \right )=\frac{3}{8}$
If she tossed the coin once and exactly 1 tail show up then total favorable.
$P\left ( \frac{A}{E_2} \right )=\frac{1}{2}$
The probability that she threw 3, 4, 5, 6 with given that she got exactly one tail or in other word
$P\left ( \frac{A}{E_2} \right )$
Using Baye’s theorem we get
$\begin{aligned} &P\left (\frac{E_2}{A} \right )=\frac{P(E_2).P\left ( \frac{A}{E_2} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )}\\ &=\frac{{\frac{2}{3}\times \frac{1}{2} }}{\frac{1}{3}\times \frac{3}{8}+\frac{2}{3}\times \frac{1}{2}}\\ &=\frac{\frac{1}{3}}{\frac{1}{8}+\frac{1}{3}}\\ &=\frac{8}{11} \end{aligned}$

Probability exercise 30.7 question 6

Answer:
$\frac{2}{9}$
Hint:
Use baye’s theorem.
Given:
Two groups are competing the computation for position of board of director of cooperation. The probability that the first and second group will win 0.6 and 0.4 respectively. Further if she first group win the probability of introducing new product is 0.7 and corresponding 0.3 if second group win, find probability of new product introduce was by second group.
Solution:
Let E = first group
F = second group
G = new product
We need to find probability that new product introduce was by second group.
i.e,
$P\left (\frac{F}{G} \right )$
So,
$\begin{aligned} &P\left (\frac{F}{G} \right )=\frac{P(F).P\left ( \frac{G}{F} \right )}{P(E)\times P\left ( \frac{G}{E} \right )+P(F)\times P\left ( \frac{G}{F} \right )}\\ \end{aligned}$
P(E) = probability of first group win = 0.6
P(F) = probability of second group win = 0.4
$\begin{aligned} &P\left (\frac{G}{E} \right )= \text { probability of new product of first group } =0.7\\ &P\left (\frac{G}{F} \right )= \text { probability of new product of second group } =0.3\\ \end{aligned}$
Putting values in Baye’s theorem we get
$\begin{aligned} &P\left (\frac{E_2}{A} \right )=\frac{0.4\times 0.3}{0.6\times 0.7+0.4\times 0.3 }\\ &=\frac{0.12}{0.42+0.12}\\ &=\frac{0.12}{0.54}\\ &=\frac{2}{9} \end{aligned}$

Probability exercise 30.7 question 7

Answer:
$\frac{2}{3}$
Hint:
Use Baye’s theorem.
Given:
Suppose 5 men out of 100 and 25 women out of 1000 are good orator. An orator chose at random. Find the probability that make person is selected. Assume equal number of men and women.
Solution:
Let A, E₁and E₂ two events that person is good orator.
Thus
$\begin{aligned} &P(E_1)=\frac{1}{2}\\ &P(E_2)=\frac{1}{2}\\ &P\left ( \frac{A}{E_1} \right )=\frac{5}{100}\\ &P\left ( \frac{A}{E_2} \right )=\frac{25}{1000}\\ \end{aligned}$
Using Baye’s theorem we get
$\begin{aligned} &P\left (\frac{E_1}{A} \right )=\frac{P(E_2).P\left ( \frac{A}{E_1} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )}\\ &P\left (\frac{E_2}{A} \right )=\frac{\frac{1}{2}\times \frac{5}{100}}{\frac{1}{2}\times \frac{5}{100}+\frac{1}{2}\times \frac{25}{1000}}\\ &=\frac{1}{1+\frac{1}{2}}\\ &=\frac{2}{3} \end{aligned}$

Probability exercise 30.7 question 8 maths

Answer:
$i)\: \frac{12}{17}, \quad ii)\: \frac{5}{17}$
Hint:
Use Baye’s theorem.
Given:
A letter is known to have come either from LONDON or CLIFTON on the envelope just two consecutive letter on are visible what is probability that letter has come from (i) LONDON (ii) CLIFTON
Solution:
Let E₁ is event that letter come from LONDON and E₂ is event that letter come from CLIFTON.
Thus
$P(E_1)=P(E_2)=\frac{1}{2}$
Let A be event come 2 consecutive letter on envelope are ON.
Then
$P\left ( \frac{A}{E_1} \right )=\frac{2}{5}$
And
$P\left ( \frac{A}{E_2} \right )=\frac{1}{6}$
Now
Using Baye’s theorem we get
$\begin{aligned} &P\left (\frac{E_1}{A} \right )=\frac{P(E_1).P\left ( \frac{A}{E_1} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )}\\ &P\left (\frac{E_1}{A} \right )=\frac{{\frac{1}{2}\times \frac{2}{5} }}{\frac{1}{2}\times \frac{2}{5}+\frac{1}{2}\times \frac{1}{6}}\\ &=\frac{12}{17} \end{aligned}$
i. Now
Using Baye’s theorem we get
$\begin{aligned} &P\left (\frac{E_2}{A} \right )=\frac{P(E_2).P\left ( \frac{A}{E_2} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )}\\ &P\left (\frac{E_2}{A} \right )=\frac{{\frac{1}{2}\times \frac{1}{6} }}{\frac{1}{2}\times \frac{2}{5}+\frac{1}{2}\times \frac{1}{6}}\\ &=\frac{\frac{1}{6}}{\frac{12+5}{30}}\\ &=\frac{1}{6}\times \frac{30}{17}\\ &=\frac{5}{17} \end{aligned}$

Probability exercise 30.7 question 9

Answer:
$\frac{3}{7}$
Hint:
Use Baye’s theorem.
Given:

In class 5% of boy and 10% of girl have 10 of more than 150. In this class 60% of students boys.
If a student is selected at random and found to have 10 of more than 150. Find probability of student is boy.
Solution:
Let E₁ is the student chose boy and E₂ is the student chose is girl.
Thus
$\begin{aligned} &P(E_1)=\frac{60}{100}\\ &P(E_2)=\frac{40}{100}\\ \end{aligned}$
Event E₁ and E₂ are mutually exclusive event A. a student have 10 more than 150.
$\begin{aligned} &P\left ( \frac{A}{E_1} \right )=P(\text { a boy student has 10 more than 150 })=\frac{5}{100}\\ &P\left ( \frac{A}{E_2} \right )=P(\text { a girl student has 10 more than 150 })=\frac{10}{100}\\ \end{aligned}$
Using Baye’s theorem we get
$\begin{aligned} &P\left (\frac{E_1}{A} \right )=\frac{P(E_1).P\left ( \frac{A}{E_1} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )}\\ &=\frac{\frac{60}{100}\times \frac{5}{100}}{\frac{60}{100}\times \frac{5}{100}+\frac{4}{100}\times \frac{10}{100}}\\ \end{aligned}$
Multiplying every term by 10000
$\begin{aligned} &=\frac{60\times 5}{60\times 5+40\times 10}\\ &=\frac{300}{700}\\ &=\frac{3}{7} \end{aligned}$

Probability exercise 30.7 question 10

Answer:
0.1
Hint:
Use Baye’s theorem.
Given:
A factor has three machines X, Y, Z producing 1000, 2000, 3000, bolts per day respectively. The machine X produce 1% defective bolts, Y produce 1.5% defective and Z produce 2% defective bolt.
Solution:
Let E₁,E₂,E₃ denote the event that machine X produce bolts, machine Y produce bolts and 2 produce bolt.
Let A be event that bolt is defective.
Total bolt = 1000 + 2000 + 3000 = 6000
$\begin{aligned} &P(E_1)=\frac{1000}{6000}=\frac{1}{6}\\ &P(E_2)=\frac{2000}{6000}=\frac{1}{3}\\ &P(E_3)=\frac{3000}{6000}=\frac{1}{2}\\ &P\left ( \frac{A}{E_1} \right )=1%=\frac{1}{100}\\ &P\left ( \frac{A}{E_2} \right )=1.5%=\frac{15}{1000}\\ &P\left ( \frac{A}{E_3} \right )=2%=\frac{2}{100}\\ \end{aligned}$
Using Baye’s theorem we get
$\begin{aligned} &P\left (\frac{E_2}{A} \right )=\frac{P(E_2).P\left ( \frac{A}{E_2} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )+P(E_3)\times P\left ( \frac{A}{E_3} \right )}\\ &=\frac{{\frac{1}{6}\times \frac{1}{100} }}{\frac{1}{6}\times \frac{1}{100}+\frac{1}{3}\times \frac{15}{1000}+\frac{1}{2}\times \frac{2}{100}}\\ &=\frac{\frac{1}{6}}{\frac{1}{6}+\frac{1}{2}+1}\\ &=\frac{1}{6}\times \frac{1+3+6}{6}\\ &=\frac{1}{10} \end{aligned}$

Probability exercise 30.7 question 11

Answer:
$i)\: \frac{3}{19}, \quad (ii)\frac{6}{19}, \quad (iii)\frac{10}{19}$
Hint:
Use baye’s theorem.
Given:
An insurance company is used 3000, 4000, 5000 trucks. The probability of accident involving a scooter, a car, a truck 0.02, 0.03, 0.04, one insured vehicle meets with accident. Find the probability of (i) Scooter, (ii) car, (iii)Truck
Solution:
Let E₁,E₂,E₃ denote the event that vehicle is (i) Scooter, (ii) car, (iii) Truck.
Let A be event that vehicle meet an accident.
It is given that 3000 scooter, 4000 car and 5000 truck.
Total vehicle = 3000 + 4000 + 5000 = 12000
$\begin{aligned} &P(E_1)=\frac{3000}{12000}=\frac{1}{4}\\ &P(E_2)=\frac{4000}{12000}=\frac{1}{3}\\ &P(E_3)=\frac{5000}{12000}=\frac{5}{12}\\ &P\left ( \frac{A}{E_1} \right )=0.02=\frac{2}{100}\\ &P\left ( \frac{A}{E_2} \right )=0.03=\frac{3}{100}\\ &P\left ( \frac{A}{E_3} \right )=0.04=\frac{4}{100}\\ \end{aligned}$
Using Baye’s theorem we get
Required probability
$\begin{aligned} &P\left (\frac{E_1}{A} \right )=\frac{P(E_1).P\left ( \frac{A}{E_1} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )+P(E_3)\times P\left ( \frac{A}{E_3} \right )}\\ &=\frac{{\frac{1}{4}\times \frac{2}{100} }}{\frac{1}{4}\times \frac{2}{100}+\frac{1}{3}\times \frac{3}{100}+\frac{5}{12}\times \frac{4}{100}}\\ &=\frac{\frac{1}{2}}{\frac{1}{2}+1+\frac{5}{3}}\\ &=\frac{\frac{1}{2}}{\frac{3+6+10}{6}}\\ &=\frac{3}{19} \end{aligned}$
Required probability
$\begin{aligned} &P\left (\frac{E_2}{A} \right )=\frac{P(E_2).P\left ( \frac{A}{E_2} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )+P(E_3)\times P\left ( \frac{A}{E_3} \right )}\\ &=\frac{{\frac{1}{3}\times \frac{3}{100} }}{\frac{1}{4}\times \frac{2}{100}+\frac{1}{3}\times \frac{3}{100}+\frac{5}{12}\times \frac{4}{100}}\\ &=\frac{1}{\frac{1}{2}+1+\frac{5}{3}}\\ &=\frac{1}{\frac{3+6+10}{6}}\\ &=\frac{6}{19} \end{aligned}$
Required probability
$\begin{aligned} &P\left (\frac{E_3}{A} \right )=\frac{P(E_3).P\left ( \frac{A}{E_3} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )+P(E_3)\times P\left ( \frac{A}{E_3} \right )}\\ &=\frac{{\frac{5}{12}\times \frac{4}{100} }}{\frac{1}{4}\times \frac{2}{100}+\frac{1}{3}\times \frac{3}{100}+\frac{5}{12}\times \frac{4}{100}}\\ &=\frac{\frac{5}{3}}{\frac{1}{2}+1+\frac{5}{3}}\\ &=\frac{\frac{5}{3}}{\frac{3+6+10}{6}}\\ &=\frac{10}{19} \end{aligned}$

Probability exercise 30.7 question 12 maths

Answer:
$\frac{1}{15}, \frac{2}{5},\frac{8 }{15}$
Hint:
Use baye’s theorem.
Given:
Suppose we have four boxes A, B, C, D.

Box	Red	White	Black
A	1	6	3
B	6	2	2
C	8	1	1
D	0	6	4

One of box has been selected at random and single marble drawn if the marble is red what is the probability that it drawn box A, B, C, D.
Solution:
Let
R → event that red marble
A → event select box A
B → event select box B
C → event select box C
D → event select box D
$\begin{aligned} &P(A)=\text { probability of box A }=\frac{1}{4}\\ &P(B)=\text { probability of box B }=\frac{1}{4}\\ &P(C)=\text { probability of box C }=\frac{1}{4}\\ &P(D)=\text { probability of box D }=\frac{1}{4}\\ &P\left ( \frac{R}{A} \right )=\frac{1}{10}\\ &P\left ( \frac{R}{B} \right )=\frac{6}{10}\\ &P\left ( \frac{R}{C} \right )=\frac{8}{10}\\ &P\left ( \frac{R}{D} \right )=\frac{0}{10}\\ \end{aligned}$
Using Baye’s theorem
$\begin{aligned} &P\left (\frac{A}{R} \right )=\frac{P(A).P\left ( \frac{R}{A} \right )}{P(A)\times P\left ( \frac{R}{A} \right )+P(B)\times P\left ( \frac{R}{B} \right )+P(C)\times P\left ( \frac{R}{C} \right )+P(D)\times P\left ( \frac{R}{D} \right )}\\ &=\frac{\frac{1}{4}\times \frac{1}{10} }{\frac{1}{4}\times \frac{1}{10}+\frac{1}{4}\times \frac{6}{10}+\frac{1}{4}\times \frac{8}{10}+\frac{1}{4}\times \frac{0}{10}}\\ &=\frac{\frac{1}{4}\times \frac{1}{10} }{\frac{1}{4}\times \frac{1}{10}(6+1+8)}\\ &=\frac{1}{15} \end{aligned}$
Using Baye’s theorem
$\begin{aligned} &P\left (\frac{B}{R} \right )=\frac{P(B).P\left ( \frac{R}{B} \right )}{P(A)\times P\left ( \frac{R}{A} \right )+P(B)\times P\left ( \frac{R}{B} \right )+P(C)\times P\left ( \frac{R}{C} \right )+P(D)\times P\left ( \frac{R}{D} \right )}\\ &=\frac{\frac{1}{4}\times \frac{6}{10} }{\frac{1}{4}\times \frac{1}{10}+\frac{1}{4}\times \frac{6}{10}+\frac{1}{4}\times \frac{8}{10}+\frac{1}{4}\times \frac{0}{10}}\\ &=\frac{\frac{1}{4}\times \frac{6}{10} }{\frac{1}{4}\times \frac{1}{10}(6+1+8)}\\ &=\frac{16}{15}\\ &=\frac{2}{5} \end{aligned}$
Using Baye’s theorem
$\begin{aligned} &P\left (\frac{C}{R} \right )=\frac{P(C).P\left ( \frac{R}{C} \right )}{P(A)\times P\left ( \frac{R}{A} \right )+P(B)\times P\left ( \frac{R}{B} \right )+P(C)\times P\left ( \frac{R}{C} \right )+P(D)\times P\left ( \frac{R}{D} \right )}\\ &=\frac{\frac{1}{4}\times \frac{8}{10} }{\frac{1}{4}\times \frac{1}{10}+\frac{1}{4}\times \frac{6}{10}+\frac{1}{4}\times \frac{8}{10}+\frac{1}{4}\times \frac{0}{10}}\\ &=\frac{\frac{1}{4}\times \frac{8}{10} }{\frac{1}{4}\times \frac{1}{10}(6+1+8)}\\ &=\frac{8}{15} \end{aligned}$

Probability exercise 30.7 question 13

Answer:
$\frac{5}{34}$
Hint:
Use Baye’s theorem.
Given:
The first operator A produce 10% defective items whereas the two other operate B, C 5% and 7% defective respectively. A is an on 50% of time, B on job 30% of time and C on job 20% of time. The defective item is produced by A.
Solution:
Let E₁,E₂,E₃ respective event of time consumed by machine A, B and c for job.
$\begin{aligned} &P(E_1)=50%=\frac{50}{100}=\frac{1}{2}\\ &P(E_2)=30%=\frac{30}{100}=\frac{3}{10}\\ &P(E_3)=20%=\frac{20}{100}=\frac{1}{5}\\ \end{aligned}$
Let X be event produce defective item
$\begin{aligned} &P\left ( \frac{X}{E_1} \right )=1%=\frac{1}{100}\\ &P\left ( \frac{X}{E_2} \right )=5%=\frac{5}{100}\\ &P\left ( \frac{X}{E_3} \right )=7%=\frac{7}{100}\\ \end{aligned}$
The probability that defective item was produce by A is given as
$\begin{aligned} &P\left ( \frac{E_1}{A} \right ) \end{aligned}$
Using Baye’s theorem we get
$\begin{aligned} &P\left (\frac{E_1}{X} \right )=\frac{P(E_1).P\left ( \frac{X}{E_1} \right )}{P(E_1)\times P\left ( \frac{X}{E_1} \right )+P(E_2)\times P\left ( \frac{X}{E_2} \right )+P(E_3)\times P\left ( \frac{X}{E_3} \right )}\\ &=\frac{{\frac{1}{2}\times \frac{1}{100} }}{\frac{1}{2}\times \frac{1}{100}+\frac{3}{10}\times \frac{5}{100}+\frac{1}{5}\times \frac{7}{100}}\\ &=\frac{\frac{1}{2}\times \frac{1}{100}}{\frac{1}{100}\left ( \frac{1}{2}+\frac{3}{2}+\frac{5}{7} \right )}\\ &=\frac{\frac{1}{2}}{\frac{17}{5}}\\ &=\frac{5}{34} \end{aligned}$

Probability exercise 30.7 question 14

Answer:
$\frac{5}{11}$
Hint:
Use Baye’s theorem.
Given:
50% of manufacture on machine A, 30% on B and 20% on C. 2% of item produced on A and 2% item produce on B are defective and 3% of these produce on C.
Solution:
Let
E₁= event that item is manufacture on A
E₂= event that item is manufacture on B
E₃= event that item is manufacture on C
Let E be event that item defective.
$\begin{aligned} &P(E_1)=\frac{50}{100}=\frac{1}{2}\\ &P(E_2)=\frac{30}{100}=\frac{3}{10}\\ &P(E_3)=\frac{20}{100}=\frac{1}{5}\\ \end{aligned}$
$\begin{aligned} &P\left ( \frac{X}{E_1} \right )=\frac{2}{100}=\frac{1}{50}\\ &P\left ( \frac{X}{E_2} \right )=\frac{2}{100}=\frac{1}{50}\\ &P\left ( \frac{X}{E_3} \right )=\frac{3}{100}\\ \end{aligned}$
Using Baye’s theorem we get
$\begin{aligned} &P\left (\frac{E_1}{X} \right )=\frac{P(E_1).P\left ( \frac{X}{E_1} \right )}{P(E_1)\times P\left ( \frac{X}{E_1} \right )+P(E_2)\times P\left ( \frac{X}{E_2} \right )+P(E_3)\times P\left ( \frac{X}{E_3} \right )}\\ &=\frac{{\frac{1}{2}\times \frac{1}{50} }}{\frac{1}{2}\times \frac{1}{50}+\frac{3}{10}\times \frac{1}{50}+\frac{1}{5}\times \frac{3}{100}}\\ &=\frac{\frac{1}{100}}{\frac{1}{100}+\frac{3}{500}+\frac{3}{500}}\\ &=\frac{\frac{1}{100}}{\frac{5+3+3}{500}}\\ &=\frac{5}{11} \end{aligned}$

Probability exercise 30.7 question 15

Answer:
$\frac{20}{43}$
Hint:
Use Baye’s theorem.
Given:
There are three coins. One is two headed coin another is biased coin that come up head 75% of time and third is also a biased coin that come up tall 40% of time. One of three coins is chosen at random and tossed and it show head. What is the probability that it was two head coins.
Solution:
E₁ be event of selecting of two headed coin
E₂ be event selecting biased coin that comes up head 75% of time.
E₃ be event selected biased coin that come up 40% of time tail.
A be the event that head comes.
Then,
$\begin{aligned} &P(E_1)=P(E_2)=P(E_3)=\frac{1}{3}\\ &P\left ( \frac{A}{E_1} \right )=\text { probability of getting head on coin is two headed }=1\\ &P\left ( \frac{A}{E_2} \right )=\text { probability of getting biased coin that come 75% of head }\\ &=\frac{75}{100}=\frac{3}{4}\\ &P\left ( \frac{A}{E_3} \right )=\text { probability of getting biased coin that come 40% of head }\\ &=\frac{40}{100}=\frac{2}{5}\\ \end{aligned}$
Using Baye’s theorem we get
$\begin{aligned} &P\left (\frac{E_1}{A} \right )=\frac{P(E_1).P\left ( \frac{A}{E_1} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )+P(E_3)\times P\left ( \frac{A}{E_3} \right )}\\ &=\frac{{\frac{1}{3}\times 1 }}{\frac{1}{3}\times 1+\frac{1}{3}\times \frac{3}{4}+\frac{1}{3}\times \frac{2}{5}}\\ &=\frac{20}{43} \end{aligned}$

Probability exercise 30.7 question 16 maths

Answer:
0, 2
Hint:
Use Baye’s theorem.
Given:
In factory machine produce 30% total item B produce 25% and C produce remaining output. If defective item produce by machine A, B, C is 1%. 12% and 2% respectively. Three machines working together produce 10,000 item in day. An item is drawn at random from day output and found defective. Find probability that was produce by machine A.
Solution:
Let E₁, E₂,E₃ event as define
E₁ =item is produce by machine A
E₂= item is produce by machine B
E₃ = item is produce by machine C
A=total event that item is defective
$\begin{aligned} &P(E_1)=\frac{30}{100}\\ &P(E_2)=\frac{25}{100}\\ &P(E_3)=\frac{45}{100}\\ &P\left ( \frac{A}{E_1} \right )=\frac{1}{100}\\ &P\left ( \frac{A}{E_2} \right )=\frac{1.2}{100}\\ &P\left ( \frac{A}{E_3} \right )=\frac{2}{100}\\ \end{aligned}$
Using Baye’s theorem we get
$\begin{aligned} &P\left (\frac{E_2}{A} \right )=\frac{P(E_2).P\left ( \frac{A}{E_2} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )+P(E_3)\times P\left ( \frac{A}{E_3} \right )}\\ &=\frac{{\frac{25}{100}\times \frac{1.2}{100} }}{\frac{30}{100}\times \frac{1}{100}+\frac{25}{100}\times \frac{1.2}{100}+\frac{45}{100}\times \frac{2}{100}}\\ &=\frac{30}{30+30+90}\\ &=\frac{1}{5}\\ &=0.2 \end{aligned}$

Probability exercise 30.7 question 18

Answer:
$\frac{36}{61}$
Hint:
Use Baye’s theorem.
Given:
Three win A, B, C contain 6 red and 4 white; 2 red, 6 white and 1 red 5 white ball respectively. A win is chosen at random and ball is drawn. If ball drawn is found to red. Find the probability that ball from win A.
Solution:
Let E₁ and E₂ denotes the event that ball is red, bag A chose B chosen and bag C chosen respectively.
$\begin{aligned} &P(E_1)=\frac{1}{3}\\ &P(E_2)=\frac{1}{3}\\ &P(E_3)=\frac{1}{3}\\ &P\left ( \frac{A}{E_1} \right )=\frac{6}{10}=\frac{3}{5}\\ &P\left ( \frac{A}{E_2} \right )=\frac{2}{8}=\frac{1}{4}\\ &P\left ( \frac{A}{E_3} \right )=\frac{1}{6}\\ \end{aligned}$
Using Baye’s theorem we get
$\begin{aligned} &P\left (\frac{E_1}{A} \right )=\frac{P(E_1).P\left ( \frac{A}{E_1} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )+P(E_3)\times P\left ( \frac{A}{E_3} \right )}\\ &=\frac{{\frac{1}{3}\times \frac{3}{5} }}{\frac{1}{3}\times \frac{3}{5}+\frac{1}{3}\times \frac{1}{4}+\frac{1}{3}\times \frac{1}{6}}\\ &=\frac{\frac{3}{5}}{\frac{3}{5}+\frac{1}{4}+\frac{1}{6}}\\ &=\frac{36}{61} \end{aligned}$

Probability exercise 30.7 question 19

Answer:
$\frac{28}{45}$
Hint:
Use Baye’s theorem.
Given:
In group of 400 people 160 smoke and non-vegetarians, 100 smoke and vegetarians. The probability of getting chest disease 35%, 20%, 10% respectively. A person chosen from group at found suffering. Find probability selected person smoke non vegetarian.
Solution:
Let us denote group of smoker and non-vegetarian as A.
B = smoker and non-vegetarian
C = no smoker and vegetarian
No of people in group A = 160;
$P(A)=\frac{160}{400}$
No of people in group B = 100;
$P(B)=\frac{100}{400}$
No of people in group C = 400 - 260 = 140;
$\begin{aligned} &P(C)=\frac{140}{400}\\ &P(D_A)=35%\\ &P(D_B)=20%\\ &P(D_C)=10%\\ \end{aligned}$
Probability of disease person from group A is
Using Baye’s theorem we get
$\begin{aligned} &P(\text { smoker and vegetarian })=\frac{P(A)\times P(D_A)}{P(A)\times P(D_A)+P(B)\times P(D_B)+P(C)\times P(D_C)}\\ &=\frac{\frac{2}{5}\times 35%}{\frac{2}{5}\times 35%+\frac{1}{4}\times 20%+\frac{7}{20}\times 10%}\\ &=\frac{\frac{2}{5}\times 35}{\frac{2}{5}\times 35+\frac{1}{4}\times 20+\frac{7}{20}\times 10}\\ &=\frac{14}{14+5+\frac{7}{2}}\\ &=\frac{14}{\frac{19+7}{2}}\\ &=\frac{28}{38+7}\\ &=\frac{28}{45} \end{aligned}$

Probability exercise 30.7 question 20 maths

Answer:
$\frac{2}{23}$
Hint:
Use Baye’s theorem.
Given:
A factory has three machine A, B, C which produce 100, 200, 300 item of particular type. The machine produce 2%, 3% and 5% defective item respectively.
Solution:
Let E₁,E₂,E₃ machine A, B and C.
D = defective item
Total production = 100 + 200 + 300 = 600
$\begin{aligned} &P(E_1)=\frac{100}{600}=\frac{1}{6}\\ &P(E_2)=\frac{200}{600}=\frac{1}{3}\\ &P(E_3)=\frac{300}{600}\\ &P\left ( \frac{D}{E_1} \right )=0.02\\ &P\left ( \frac{D}{E_2} \right )=0.03\\ &P\left ( \frac{D}{E_3} \right )=0.05\\ \end{aligned}$
Using Baye’s theorem we get
$\begin{aligned} &P\left (\frac{E_1}{D} \right )=\frac{P(E_1).P\left ( \frac{D}{E_1} \right )}{P(E_1)\times P\left ( \frac{D}{E_1} \right )+P(E_2)\times P\left ( \frac{D}{E_2} \right )+P(E_3)\times P\left ( \frac{D}{E_3} \right )}\\ &=\frac{{\frac{1}{6}\times 0.02 }}{\frac{1}{6}\times 0.02+\frac{1}{3}\times 0.03+\frac{1}{2}\times 0.05}\\ &=\frac{2}{23} \end{aligned}$

Probability exercise 30.7 question 21

Answer:
$\frac{1}{5}$
Hint:
Use Baye’s theorem.
Given:
A bag contain 1 white and 6 red ball, and second bag contain 4 white 3 red. One of the bags is pick at random and ball is randomly drawn from college to white in color.
Solution:
Let E₁, E₂,E₃ be event
E₁ = E₂ = Selection of bag,
A=The white ball drawn random
$\begin{aligned} &P(E_1)=\frac{1}{2}\\ &P(E_2)=\frac{1}{2}\\ &P\left ( \frac{A}{E_1} \right )=\frac{1}{7}\\ &P\left ( \frac{A}{E_2} \right )=\frac{4}{7}\\ \end{aligned}$
Using Baye’s theorem we get
$\begin{aligned} &P\left (\frac{E_1}{A} \right )=\frac{P(E_1).P\left ( \frac{A}{E_1} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )}\\ &=\frac{{\frac{1}{2}\times \frac{1}{7} }}{\frac{1}{2}\times \frac{1}{7}+\frac{1}{2}\times \frac{4}{7}}\\ &=\frac{1}{1+4}\\ &=\frac{1}{5} \end{aligned}$

Probability exercise 30.7 question 22

Answer:
$\frac{3}{11}$
Hint:
Use Baye’s theorem.
Given:
In certain college 4% of boy and 1% of girl are taller than 1.75 meter. Furthermore 60% of student in colleges are girl. A student selected at garden from college is found be taller 1.75 meter.
Solution:
Let E₁ and E₂ denote the selected student is a girl and selected student is a boy.
A denoted the student is 1.75m tall.

$\begin{aligned} &P(E_1)=\frac{60}{100}\\ &P(E_2)=\frac{40}{100}\\ &P\left ( \frac{A}{E_1} \right )=\frac{1}{100}\\ &P\left ( \frac{A}{E_2} \right )=\frac{4}{100}\\ \end{aligned}$
Using Baye’s theorem we get

$\begin{aligned} &P\left (\frac{E_1}{A} \right )=\frac{P(E_1).P\left ( \frac{A}{E_1} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )}\\ &=\frac{{\frac{60}{100}\times \frac{1}{100} }}{\frac{60}{100}\times \frac{1}{100}+\frac{40}{100}\times \frac{4}{100}}\\ &=\frac{6}{6+16}\\ &=\frac{6}{22}\\ &=\frac{3}{11} \end{aligned}$

Probability exercise 30.7 question 24 maths

Answer:
$\frac{7}{10}$
Hint:
Use Baye’s theorem.
Given:
The person A, B, C applies for job of manager in private company. Chance of their selection is in ratio 1:2:4. The probability that A, B, C can introduce change to improve profit of company are 0.8, 0.5, 0.3 respectively.
Solution:
Let E₁ ,E₂,E₃ be event that the selection of A, B, C.
$\begin{aligned} &P(E_1)=\text { Probability of selection of A }=\frac{1}{7}\\ &P(E_2)=\text { Probability of selection of B }=\frac{2}{7}\\ &P(E_3)=\text { Probability of selection of C }=\frac{4}{7}\\ &P\left ( \frac{A}{E_1} \right )=\text { Probability that A doesnot introduce }=0.2\\ &P\left ( \frac{A}{E_2} \right )\text { Probability that B doesnot introduce }=0.5\\ &P\left ( \frac{A}{E_3} \right )=\text { Probability that C doesnot introduce }=0.7\\ \end{aligned}$
Using Baye’s theorem we get
$\begin{aligned} &P\left (\frac{E_3}{A} \right )=\frac{P(E_3).P\left ( \frac{A}{E_3} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )+P(E_3)\times P\left ( \frac{A}{E_3} \right )}\\ &=\frac{{\frac{4}{7}\times 0.7}}{\frac{1}{7}\times 0.2+\frac{2}{7}\times 0.5+\frac{4}{7}\times 0.7}\\ &=\frac{28}{0.2+1+2.8}\\ &=\frac{28}{4}\\ &=\frac{7}{10} \end{aligned}$

Probability exercise 30.7 question 25

Answer:
$\frac{3}{4}$
Hint:
Use Baye’s theorem.
Given:
An insurance company insures 2000 scooter and 3000 motorcycle. The probability of an accident involve scooter is 0.01 and that of motorcycle.
Solution:
$\begin{aligned} &P(E_1)=\frac{2000}{5000}=0.4\\ &P(E_2)=\frac{3000}{5000}=0.6\\ &P\left ( \frac{A}{E_1} \right )=0.01\\ &P\left ( \frac{A}{E_2} \right )=0.02\\ \end{aligned}$
Using Baye’s theorem we get
$\begin{aligned} &P\left (\frac{E_2}{A} \right )=\frac{P(E_2).P\left ( \frac{A}{E_2} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )}\\ &=\frac{{0.6\times 0.02}}{0.4\times 0.01+0.6\times 0.02}\\ &=\frac{0.012}{0.004+0.012}\\ &=\frac{3}{4} \end{aligned}$

Probability exercise 30.7 question 26

Answer:
$\frac{9}{13}$
Hint:
Use Baye’s theorem.
Given:
It is known that 60%beside in a hostel and 40% don’t reside in hostel. Previous year results report that 30% of student residing in hostel attain. A grade and 20% of ones not residing in hostel attain.
Solution:
Let A denotes A -grade
H denote from hostel
D denote day scholar
$\begin{aligned} &P(H)=0.6\\ &P(D)=0.4\\ &P\left ( \frac{A}{H} \right )=0.3\\ &P\left ( \frac{A}{D} \right )=0.2\\ \end{aligned}$
Using Baye’s theorem we get
$\begin{aligned} &P\left (\frac{H}{A} \right )=\frac{P(H).P\left ( \frac{A}{H} \right )}{P(H)\times P\left ( \frac{A}{H} \right )+P(D)\times P\left ( \frac{A}{D} \right )}\\ &=\frac{{0.3\times 0.6 }}{0.3\times 0.6+0.2\times 0.4}\\ &=\frac{0.18}{0.08}\\ &=\frac{18}{26}\\ &=\frac{9}{13} \end{aligned}$

Probability exercise 30.7 question 27

Answer:
$\frac{4}{9}$
Hint:
Use Baye’s theorem.
Given:
There are three coins one is two headed coin, another is blazed coin that comes up heads 75% of time and third is unbiased coin.
Solution:
Let
C₁= two headed coin
C₂= biased coin
C₃= unbiased coin
We need to find probability that coin is two headed it show
$P\left (\frac{C_1}{H} \right )$
$\begin{aligned} &P(C_1)=P(C_2)=P(C_3)=\frac{1}{3}\\ &P\left ( \frac{A}{C_1} \right )=1\\ &P\left ( \frac{A}{C_2} \right )=\frac{3}{4}\\ &P\left ( \frac{A}{C_3} \right )=\frac{1}{2}\\ \end{aligned}$
Using Baye’s theorem we get
$\begin{aligned} &P\left (\frac{C_1}{H} \right )=\frac{P(C_1).P\left ( \frac{H}{C_1} \right )}{P(C_1)\times P\left ( \frac{H}{C_1} \right )+P(C_2)\times P\left ( \frac{H}{C_2} \right )+P(C_3)\times P\left ( \frac{H}{C_3} \right )}\\ &=\frac{{\frac{1}{3}\times 1 }}{\frac{1}{3}\times 1+\frac{1}{3}\times \frac{3}{4}+\frac{1}{3}\times \frac{1}{2}}\\ &=\frac{\frac{1}{3}\times 1}{\frac{1}{3}\left ( 1+\frac{3}{4}+\frac{1}{2} \right )}\\ &=\frac{1}{\frac{9}{4}}\\ &=\frac{4}{9} \end{aligned}$

Probability exercise 30.7 question 28 maths

Answer:
$\frac{14}{29}$
Hint:
Use Baye’s theorem.
Given:
Assume that chance of patient having a heart attack is 40%. If is also assumed that meditation and yoga source reduce the rest of heart attack by 30% and preparation of certain drug reduce its chance by 25%.
Solution:
Let A₁,E₁ and E₂ denote event that person has heart attack, the selected person followed the source of yoga and meditation and person adopt drug precaution.
$\begin{aligned} &P(A)=0.40\\ &P(E_1)=P(E_2)=\frac{1}{2}\\ &P\left ( \frac{A}{E_1} \right )=0.28=(0.40\times 0.70)\\ &P\left ( \frac{A}{E_2} \right )=0.40\times 0.75=0.30\\ \end{aligned}$
Using Baye’s theorem we get
$\begin{aligned} &P\left (\frac{E_1}{A} \right )=\frac{P(E_1).P\left ( \frac{A}{E_1} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )}\\ &=\frac{{\frac{1}{2}\times 0.28 }}{\frac{1}{2}\times 0.28+\frac{1}{2}\times 0.30}\\ &=\frac{14}{29} \end{aligned}$

Probability exercise 30.7 question 29

Answer:
$\frac{156}{947}$
Hint:
Use baye’s theorem.
Given:

Box	Colour
Box	Black	White	Red	Blue
I	3	4	5	6
II	2	2	2	2
III	1	2	3	1
IV	4	3	1	5

A box is selected at random and ball is randomly drawn from the selected box. The color the ball is black. What is the probability that ball drawn is from the box III.
Solution:
Let A: Event that is black ball is selected
E₁ : Event that the ball i selecting from box I
E₂ : Event that the ball is selected for box II
E₃ : Event that the ball is selected for box III
E₄ : Event that the ball is selected for box IV
$\begin{aligned} &P(E_1)=\text { Probability that ball drawn is from box I }=\frac{1}{4}\\ &P(E_2)=\text { Probability that ball drawn is from box II }=\frac{1}{4}\\ &P(E_3)=\text { Probability that ball drawn is from box III }=\frac{1}{4}\\ &P(E_4)=\text { Probability that ball drawn is from box IV }=\frac{1}{4}\\ &P\left ( \frac{A}{E_1} \right )=\frac{\text { no.of balls }}{\text { Total no.of balls in the box }}=\frac{3}{18}\\ &P\left ( \frac{A}{E_2} \right )=\frac{\text { no.of balls }}{\text { Total no.of balls in the box }}=\frac{2}{8}=\frac{1}{4}\\ &P\left ( \frac{A}{E_3} \right )=\frac{1}{7}\\ &P\left ( \frac{A}{E_4} \right )=\frac{4}{13}\\ \end{aligned}$
Using Baye’s theorem we get
$\begin{aligned} &P\left (\frac{E_3}{A} \right )=\frac{P(E_3).P\left ( \frac{A}{E_3} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )+P(E_3)\times P\left ( \frac{A}{E_3} \right )+P(E_4)\times P\left ( \frac{A}{E_4} \right )}\\ &=\frac{{\frac{1}{4}\times \frac{1}{7} }}{\frac{1}{4}\times \frac{1}{6}+\frac{1}{4}\times \frac{1}{4}+\frac{1}{4}\times \frac{1}{7}+\frac{1}{4}\times \frac{4}{13}}\\ &=\frac{\frac{1}{28}}{\frac{1}{24}+\frac{1}{16}+\frac{1}{28}+\frac{1}{13}}\\ &=\frac{156}{611+336}\\ &=\frac{156}{947} \end{aligned}$

Probability exercise 30.7 question 30

Answer:
$\frac{81}{85}$
Hint:
Use Baye’s theorem.
Given:
If machine is correctly set up it produce 90% acceptable item. If it incorrectly set up the produce 40% acceptable item. Past experience show 80% of setup is correctly done. If after certain set up the machine produce 2 acceptance items.
Solution:
Let A be event that machine produce 2 acceptable item.
E₁ Represents event of correct setup.
E₂ Represents event of incorrect setup.
$\begin{aligned} &P(E_1)=0.8\\ &P(E_2)=0.2\\ &P\left ( \frac{A}{E_1} \right )=0.9\times 0.9\\ &P\left ( \frac{A}{E_2} \right )=0.4\times 0.4\\ \end{aligned}$
Using Baye’s theorem we get
$\begin{aligned} &P\left (\frac{E_2}{A} \right )=\frac{P(E_2).P\left ( \frac{A}{E_2} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )}\\ &=\frac{{\frac{8}{10}\times \frac{81}{100} }}{\frac{8}{10}\times \frac{81}{100}+\frac{2}{10}\times \frac{16}{100}}\\ &=\frac{648}{680}\\ &=\frac{81}{85} \end{aligned}$

Probability exercise 30.7 question 31

Answer:
$\frac{18}{133}$
Hint:
Use Baye’s theorem.
Given:
Bag A contain 3 red, 5 black, white bag contain 4 red and 4 black. Two ball are transferred at random from bag A to bag B and then a ball B drawn from bag B at random..
Solution:
2 ball drawn from bag A could be both red probability =
$\frac{3C_2}{8C_2}$
2 one red and one black probability =
$\frac{3\times 5}{8C_2}$
3 both black probability =
$\frac{8C_2}{5C_2}$
The no of ball in bag B in each case would be
1. 6 red 4 black = probability pick red =
$\frac{6}{10}$
2. 5 red 5 black = probability pick red =
$\frac{5}{10}$
3. 4 red 6 black = probability pick red =
$\frac{4}{10}$
Probability of two red ball transfer under the condition that red ball found
Using Baye’s theorem we get
$=\frac{\frac{3C_2}{8C_2}\times \frac{6}{10}}{\frac{3C_2}{8C_2}\times \frac{6}{10}+\frac{3\times 5}{8C_2}\times \frac{5}{10}+\frac{3C_2}{5C_2}\times \frac{4}{10}}\\$
$=\frac{18}{133}$

Probability exercise 30.7 question 32 maths

Answer:
$\frac{110}{221}$
Hint:
Use Baye’s theorem.
Given:
Probability that T.B is detected when a person is actually suffering is 0.99. The probability that doctor diagnose incorrectly that person has T.B on basis of x-ray is 0.001. In certain city 1 in 1000 person suffer from T.B.
Solution:
Let
E₁= Event that person has T.B
E₂=Event that person does not have T.B
E=Event that person is diagnose to have T.B
$\begin{aligned} &P(E_1)=\frac{1}{1000}=0.001\\ &P(E_2)=\frac{999}{1000}=0.999\\ &P\left ( \frac{E}{E_1} \right )=0.99\\ &P\left ( \frac{E}{E_2} \right )=0.001\\ \end{aligned}$
And
Using Baye’s theorem we get
$\begin{aligned} &P\left (\frac{E_1}{E} \right )=\frac{P(E_1).P\left ( \frac{E}{E_1} \right )}{P(E_1)\times P\left ( \frac{E}{E_1} \right )+P(E_2)\times P\left ( \frac{E}{E_2} \right )}\\ &=\frac{0.001\times 0.99}{0.001\times 0.99+0.999\times 0.001} \\ &=\frac{990}{990+999}\\ &=\frac{110}{110+111}\\ &=\frac{110}{221} \end{aligned}$

Probability exercise 30.7 question 33

Answer:
$\frac{90}{589}$
Hint:
Use Baye’s theorem.
Given:
The test will correctly detect the disease 1% of time probability large population of which an estimate 0.2% has disease a person.
Solution:
Let A,E₁ and E₂ denote the event that the person suffer from the disease, the test detecte the disease correctly and the test does not detect the disease correctly, respectively.
$\begin{aligned} &P(E_1)=0.002\\ &P(E_2)=0.998\\ \end{aligned}$
Now
$\begin{aligned} &P\left ( \frac{A}{E_1} \right )=0.90\\ &P\left ( \frac{A}{E_2} \right )=0.01\\ \end{aligned}$
Using Baye’s theorem we get
$\begin{aligned} &P\left (\frac{E_1}{A} \right )=\frac{P(E_1).P\left ( \frac{A}{E_1} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )}\\ &=\frac{{\frac{90}{100}\times \frac{2}{1000} }}{\frac{90}{100}\times \frac{2}{1000}+\frac{1}{100}\times \frac{998}{1000}}\\ &=\frac{180}{180+998}\\ &=\frac{180}{1178}\\ &=\frac{90}{589} \end{aligned}$

Probability exercise 30.7 question 34

Answer:
D₁
Hint:
Use baye’s theorem.
Given:
1800 had disease d₁ , 2100 has disease d₂ , and the other disease d₃.1500 patient with disease d₁, 1200 disease d₃ is 900.
Solution:
Let A, E₁,E₂,E₃ denote the events that the patient show symptom patient has disease d₁ , has disease d₂ and has disease d₃ , respectively.
$\begin{aligned} &P(E_1)=\frac{1800}{5000}\\ &P(E_2)=\frac{2100}{5000}\\ &P(E_3)=\frac{1100}{5000}\\ \end{aligned}$
Now,
$\begin{aligned} &P\left ( \frac{A}{E_1} \right )=\frac{1500}{1800}\\ &P\left ( \frac{A}{E_2} \right )=\frac{1200}{2100}\\ &P\left ( \frac{A}{E_3} \right )=\frac{9000}{1100}\\ \end{aligned}$
Using Baye’s theorem we get
Required probability
$\begin{aligned} &P\left (\frac{E_1}{A} \right )=\frac{P(E_1).P\left ( \frac{A}{E_1} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )+P(E_3)\times P\left ( \frac{A}{E_3} \right )}\\ &=\frac{{\frac{1800}{5000}\times \frac{1500}{1800} }}{\frac{1800}{5000}\times \frac{1500}{1800}+\frac{2100}{5000}\times \frac{1200}{2100}+\frac{1100}{5000}\times \frac{9000}{1100}}\\ &=\frac{15}{15+12+9}\\ &=\frac{15}{12} \end{aligned}$
Required probability

$\begin{aligned} &P\left (\frac{E_2}{A} \right )=\frac{P(E_2).P\left ( \frac{A}{E_2} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )+P(E_3)\times P\left ( \frac{A}{E_3} \right )}\\ &=\frac{{\frac{2100}{5000}\times \frac{1200}{2100} }}{\frac{1800}{5000}\times \frac{1500}{1800}+\frac{2100}{5000}\times \frac{1200}{2100}+\frac{1100}{5000}\times \frac{9000}{1100}}\\ &=\frac{12}{15+12+9}\\ &=\frac{12}{36} \end{aligned}$
Required probability
$\begin{aligned} &P\left (\frac{E_3}{A} \right )=\frac{P(E_3).P\left ( \frac{A}{E_3} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )+P(E_3)\times P\left ( \frac{A}{E_3} \right )}\\ &=\frac{{\frac{1100}{5000}\times \frac{9000}{1100} }}{\frac{1800}{5000}\times \frac{1500}{1800}+\frac{2100}{5000}\times \frac{1200}{2100}+\frac{1100}{5000}\times \frac{9000}{1100}}\\ &=\frac{9}{15+12+9}\\ &=\frac{9}{36}\\ &=\frac{1}{4} \end{aligned}$
As
$\begin{aligned} &P\left (\frac{E_1}{A} \right ) \end{aligned}$
is maximum, so, it is most likely that the person suffer from the disease d₁.

Probability exercise 30.7 question 35

Answer:
$\frac{3}{13}$
Hint:
Use baye’s theorem.
Given:
A is known to speak with 3 time out of 5 time. He throws a die and report that is one. Find the probability that actually one.
Solution:
Let A, E₁ and E₂ denote the events that the man reports the appearance of 1 on throwing a die, 1 occur and 1 does not occur respectively.
$\begin{aligned} &P(E_1)=\frac{1}{6}\\ &P(E_2)=\frac{5}{6}\\ \end{aligned}$
Now,
$\begin{aligned} &P\left ( \frac{A}{E_1} \right )=\frac{3}{5}\\ &P\left ( \frac{A}{E_2} \right )=\frac{2}{5}\\ \end{aligned}$
Using Baye’s theorem we get
Required probability
$\begin{aligned} &P\left (\frac{E_1}{A} \right )=\frac{P(E_1).P\left ( \frac{A}{E_1} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )}\\ &=\frac{{\frac{1}{6}\times \frac{3}{5} }}{\frac{1}{6}\times \frac{3}{5}+\frac{5}{6}\times \frac{2}{5}}\\ &=\frac{3}{3+10}\\ &=\frac{3}{13} \end{aligned}$

Probability exercise 30.7 question 36 maths

Answer:
$\frac{4}{9}$
Hint:
Use baye’s theorem.
Given:
A speaks the truth 8 time out of 10 time a die is tossed. He report that it was 5.
Solution:
Let A denote the event that man reports 5 occur and E the event that actually 5 tossed up.
$\begin{aligned} &P(E)=\frac{1}{6}\\ &P(E)=1-\frac{1}{6}=\frac{5}{6}\\ \end{aligned}$
Also,
$\begin{aligned} &P\left ( \frac{A}{E} \right ) \end{aligned}$
=Probability that man reports that 5 occur given that 5 actually turned up
= probability of man speak the truth
$\begin{aligned} &=\frac{8}{10}=\frac{4}{5} \end{aligned}$
Probability that man reports that 5 occur given that 5 does not turned up
= probability of man not speak the truth
$\begin{aligned} &=1-\frac{4}{5}=\frac{1}{5} \end{aligned}$
Using Baye’s theorem we get
Required probability
$\begin{aligned} &P\left (\frac{E}{A} \right )=\frac{P(E_).P\left ( \frac{A}{E} \right )}{P(E)\times P\left ( \frac{A}{E} \right )+P(E)\times P\left ( \frac{A}{E} \right )}\\ &=\frac{\frac{1}{6}\times \frac{4}{5}}{\frac{1}{6}\times \frac{4}{5}+\frac{5}{6}\times \frac{1}{5}} \\ &=\frac{4}{9} \end{aligned}$

Probability exercise 30.7 question 37

Answer:
$\frac{12}{13}$
Hint:
Use baye’s theorem.
Given:
In answering a question on multiple choice tests, students either answers or guess. Assume the student who guess at answer will correct with probability
$\frac{1}{4}$
Solution:
Let
A = student know answer
B= student guess
C = student answer correctly
We know to find probability that student know answer if he answer is correctly.
$\begin{aligned} &P(A)=\frac{3}{4}\\ &P(B)=\frac{1}{4}\\ &P\left ( \frac{C}{A} \right )=1\\ &P\left ( \frac{C}{B} \right )=\frac{1}{4}\\ \end{aligned}$
Using Baye’s theorem we get
Required probability
$\begin{aligned} &P\left (\frac{A}{C} \right )=\frac{P(A).P\left ( \frac{C}{A} \right )}{P(A)\times P\left ( \frac{C}{A} \right )+P(B)\times P\left ( \frac{C}{B} \right )}\\ &=\frac{{\frac{3}{4}\times 1 }}{\frac{3}{4}\times 1+\frac{1}{4}\times \frac{1}{4}}\\ &=\frac{\frac{1}{4}\times 1}{\frac{1}{4}\left ( \frac{1}{4}+3 \right )}\\ &=\frac{3}{\frac{1}{4}+3}\\ &=\frac{3}{\frac{13}{4}}\\ &=\frac{12}{13} \end{aligned}$

Probability exercise 30.7 question 38

Answer:
$\frac{22}{133}$
Hint:
Use Baye’s theorem.
Given:
A lab blood test 99% effective in detecting certain when its infection is present. However, the test yield a false positive result for 0.5% of healthy person 0.01% of population actually has disease.
Solution:
Let E₁ , E₂ be respective event that a person has disease and person not have disease.
Since, E₁ , E₂ are events complimentary to each other
$\begin{aligned} &P(E_1)+P(E_2)=1\\ &P(E_2)=1-0.001=0.999 \end{aligned}$
Let A be event that blood test positive.
$\begin{aligned} &P(E_1)=1%=\frac{0.1}{100}=0.001\\ &P\left ( \frac{A}{E_1} \right )=99%=0.99\\ &P\left ( \frac{A}{E_2} \right )=\text { result positive no disease }=0.5%=0.005\\ \end{aligned}$
Probability that a person have disease given that his result positive
$\begin{aligned} &P\left ( \frac{E_1}{A} \right )\ \end{aligned}$
Using Baye’s theorem we get
$\begin{aligned} &P\left (\frac{E_1}{A} \right )=\frac{P(E_1).P\left ( \frac{A}{E_1} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )}\\ &=\frac{0.001\times 0.99}{0.001\times 0.99+0.999\times 0.005}\\ &=\frac{0.00099}{0.00099+0.004995}\\ &=\frac{0.00099}{0.005985}\\ &=\frac{990}{5989}\\ &=\frac{110}{665}\\ &=\frac{22}{133} \end{aligned}$

Probability exercise 30.7 question 39

Answer:
$\frac{200}{231}$
Hint:
Use Baye’s theorem.
Given:
There are three categories of student in class 60 student.
A very hardworking, B = regular but not hardworking, C = careless and irregular, 10 student are category A, 30 in B, rest in C. if found that probability of student of category A unable to get good marks in final year, probability of student in exam is 0.002 of category B is 0.02 and category C is 0.20.
Solution:
Let
E₁= Student select category A
E₂= Student select category B
E₃= Student select category C
S=Student not get good marks
$\begin{aligned} &P(E_1)=\frac{1}{6}\\ &P(E_2)=\frac{3}{6}\\ &P(E_3)=\frac{2}{6}\\ &P\left ( \frac{S}{E_1} \right )=0.002\\ &P\left ( \frac{S}{E_2} \right )=0.002\\ &P\left ( \frac{S}{E_3} \right )=0.2\\ \end{aligned}$
Using Baye’s theorem we get
Required probability
$\begin{aligned} &P\left (\frac{E_3}{A} \right )=\frac{P(E_3).P\left ( \frac{S}{E_3} \right )}{P(E_1)\times P\left ( \frac{S}{E_1} \right )+P(E_2)\times P\left ( \frac{S}{E_2} \right )+P(E_3)\times P\left ( \frac{S}{E_3} \right )}\\ &=\frac{{\frac{2}{6}\times 0.2 }}{\frac{1}{6}\times 0.002+\frac{3}{6}\times 0.02+\frac{2}{6}\times 0.2}\\ &=\frac{200}{231} \end{aligned}$

Probability exercise 30.7 question 1

Answer:
33/118, 55/118, 30/118
Hint:
Use baye’s theorem.
Given:
The content of win I, II, III is as follows
Win I; 1 white, 2 black and 3 red balls.
Win II: 1 black, 2 white and 1 red balls,
Win III: 4 white, 5 black, 3 red balls
One win is chosen at random and two balls are drawn. The happen to be white and red what is probabilities that come from win I, II, III.
Solution:
Let E₁,E₂,E₃ be selecting win I, win II, and win III.
Let A be event of chosen two ball (w,R).
$\begin{aligned} &P(E_1)=\frac{1}{3}\\ &P(E_2)=\frac{1}{3}\\ &P(E_3)=\frac{1}{3}\\ &P\left (\frac{A}{E_1} \right )=\frac{1C_1\times 3C_1}{6C_2}\\ &=\frac{1\times 3}{15}\\ &=\frac{1}{5}\\ &P\left (\frac{A}{E_2} \right )=\frac{2C_1\times 1C_1}{4C_2}\\ &=\frac{2\times 1}{6}\\ &=\frac{1}{3}\\ \end{aligned}$
$\begin{aligned} &P\left (\frac{A}{E_3} \right )=\frac{4C_1\times 3C_1}{12C_2}\\ &=\frac{4\times 3}{66}\\ &=\frac{2}{11}\\ \end{aligned}$
Using Bayes’ theorem we get
Required probability
$\begin{aligned} &P\left (\frac{E_1}{A} \right )=\frac{P(E_1).P\left ( \frac{A}{E_1} \right )}{P(E_1).P\left ( \frac{A}{E_1} \right )+P(E_2).P\left ( \frac{A}{E_2} \right )+P(E_3).P\left ( \frac{A}{E_3} \right )}\\ &=\frac{{\frac{1}{3}.\left ( \frac{1}{5} \right )}}{\frac{1}{3}\times \frac{1}{5}+\frac{1}{3}\times \frac{1}{3}+\frac{1}{3}\times \frac{2}{11}}\\ &=\frac{\frac{1}{5}}{\frac{37+55+30}{165}}\\ &=\frac{1}{5}\times \frac{165}{118}\\ &=\frac{33}{118} \end{aligned}$
Required probability
$\begin{aligned} &P\left (\frac{E_2}{A} \right )=\frac{P(E_2).P\left ( \frac{A}{E_2} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )+P(E_3)\times P\left ( \frac{A}{E_3} \right )}\\ &=\frac{{\frac{1}{3}.\left ( \frac{1}{3} \right )}}{\frac{1}{3}\times \frac{1}{5}+\frac{1}{3}\times \frac{1}{3}+\frac{1}{3}\times \frac{2}{11}}\\ &=\frac{\frac{1}{3}}{\frac{37+55+30}{165}}\\ &=\frac{1}{3}\times \frac{165}{118}\\ &=\frac{55}{118} \end{aligned}$
Required probability
$\begin{aligned} &P\left (\frac{E_3}{A} \right )=\frac{P(E_3).P\left ( \frac{A}{E_3} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )+P(E_3)\times P\left ( \frac{A}{E_3} \right )}\\ &=\frac{{\frac{1}{3}.\left ( \frac{2}{11} \right )}}{\frac{1}{3}\times \frac{1}{5}+\frac{1}{3}\times \frac{1}{3}+\frac{1}{3}\times \frac{2}{11}}\\ &=\frac{\frac{2}{11}}{\frac{37+55+30}{165}}\\ &=\frac{2}{11}\times \frac{165}{118}\\ &=\frac{30}{118} \end{aligned}$

Probability exercise 30.7 question 4 maths

Answer:
$\begin{aligned} &\frac{1}{40} \end{aligned}$
Hint:
Use baye’s theorem.
Given:
The content of three win
Win I: 7 white, 3 black;
Win II: 4 white, 6 black
Win III: 2 white 8 black
One ball drawn as random with probability 0.20, 0.60, 0.20 respectively drawn chosen win two balls drawn random without replacement. If both these ball drawn what is probability came from win III?
Solution:
Let E₁,E₂,E₃ denote the event of selecting win I, win II, win III.
Let A be event of two balls drawn white.
$\begin{aligned} &P(E_1)=\frac{20}{100}\\ &P(E_2)=\frac{60}{100}\\ &P(E_3)=\frac{20}{100}\\ &P\left ( \frac{A}{E_1} \right )=\frac{7C_2}{10C_2}=\frac{21}{45}\\ &P\left ( \frac{A}{E_2} \right )=\frac{4C_2}{10C_2}=\frac{6}{45}\\ &P\left ( \frac{A}{E_3} \right )=\frac{2C_2}{10C_2}=\frac{1}{45}\\ \end{aligned}$
Using Baye’s theorem we get
$\begin{aligned} &P\left (\frac{E_3}{A} \right )=\frac{P(E_3).P\left ( \frac{A}{E_3} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )+P(E_3)\times P\left ( \frac{A}{E_3} \right )}\\ &=\frac{{\frac{20}{100}\times \frac{1}{45} }}{\frac{20}{100}\times \frac{21}{45}+\frac{60}{100}\times \frac{6}{45}+\frac{20}{100}\times \frac{1}{45}}\\ &=\frac{1}{21+18+1}\\ &=\frac{1}{40} \end{aligned}$

Probability exercise 30.7 question 17

Answer:
$\frac{3}{7}$
Hint:
Use Bayes theorem.
Given:
The first plant manufacture 60% of bicycle and second plant 40% out of that 80% of bicycle are rated of standard quality at first plant and 90% of standard quality at second plant.
Solution:
E₁ be event of cycle is from first plant
E₂ be event of second plant
A be event the cycle is standard
$\begin{aligned} &P(E_1)=0.6\\ &P(E_2)=0.4\\ &P\left ( \frac{A}{E_1} \right )=0.8\\ &P\left ( \frac{A}{E_2} \right )=0.9\\ \end{aligned}$
Using Baye’s theorem we get
$\begin{aligned} &P\left (\frac{E_2}{A} \right )=\frac{P(E_2).P\left ( \frac{A}{E_2} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )}\\ &=\frac{0.4\times 0.9}{0.6\times 0.8+0.4\times 0.9}\\ &=\frac{0.36}{0.84}\\ &=\frac{36}{84}\\ &=\frac{9}{21}=\frac{3}{7} \end{aligned}$

Probability exercise 30.7 question 23

Answer:
$\frac{3}{5}$
Hint:
Use Baye’s theorem.
Given:
From A, B, C chance of being selected as manager of firm in ratio 4:1:2 respectively. The respectively probability for them a radical change in marketing strategy, 0.3, 0.8, 0.5.
Solution:
Let E₁ ,E₂,E₃ denote that event that the change take place is selected B is selected C is selected.
$\begin{aligned} &P(E_1)=\frac{4}{7}\\ &P(E_2)=\frac{1}{37}\\ &P(E_3)=\frac{2}{7}\\ &P\left ( \frac{A}{E_1} \right )=0.3\\ &P\left ( \frac{A}{E_2} \right )=0.8\\ &P\left ( \frac{A}{E_3} \right )=0.5\\ \end{aligned}$
Using Baye’s theorem we get
$\begin{aligned} &P\left (\frac{E_1}{A} \right )=\frac{P(E_1).P\left ( \frac{A}{E_1} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )}\\ &=\frac{{\frac{4}{7}\times 0.3 }}{\frac{4}{7}\times 0.3+\frac{1}{7}\times 0.8+\frac{2}{7}\times 0.5}\\ &=\frac{1.2}{1.2+0.8+1}\\ &=\frac{1.2}{3}\\ &=\frac{12}{30}\\ &=\frac{2}{5}\\ &\text { Required probability }=1-P\left (\frac{E_1}{A} \right )=1-\frac{2}{5}=\frac{3}{5} \end{aligned}$

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The concepts that are covered in chapter 30 of the RD Sharma class 12 solutions Probability 30.7 are conditional probability, multiplication rule, random variables, Bayes theorem, Bernoulli Trials, Binomial Distribution.

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