RD Sharma Class 12 Exercise 30.6 Probability Solutions Maths

RD Sharma Class 12 Exercise 30.6 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 30.6 solution is a pretty popular choice for NCERT solutions as deduced from student reviews. We all know that the CBSE syllabus requires a thorough understanding of the NCERT book and the questions given in the book so that they can understand the concepts clearly. Mathematics is a challenging subject that needs to be practised regularly for perfection. The RD Sharma class 12 chapter 30 exercise 30.6 will help the students to master mathematics and ace their exams. It will be the perfect book for last-minute revisions.

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RD Sharma Class 12 Solutions Chapter30 Probability - Other Exercise
Probability Excercise:30.6
RD Sharma Chapter-wise Solutions

Also Read - RD Sharma Solutions For Class 9 to 12 Maths

RD Sharma Class 12 Solutions Chapter30 Probability - Other Exercise

Probability Excercise:30.6

Probability exercise 30.6 question 1

Answer:
$\frac{39}{88}$
Hint:
We use the probability formula
Given:
A bag A contains 5 white and 6 black balls. Another bag B contains 4 white and 3 black balls.
Solution:
Let
E₁=A white ball is transferred bag A to bag B,then drawing the black ball
E₂=A black ball is transferred bag A to bag B,then drawing the black ball
A = Ball draw is back
$\begin{aligned} Probability of an event = \frac{No. of outcomes}{Total outcomes} \\ P (E_{1}) = \frac{5}{11}; P (E_{2}) = \frac{6}{11} \\ P (\frac{A}{E_{1}}) = \frac{3}{8}; P (\frac{A}{E_{2}}) = \frac{4}{8} \end{aligned}$
Using total probability theorem
$\begin{aligned} required probability = P (A) = P (E_{1}) \times P (\frac{A}{E_{1}}) + P (E_{2}) \times P (\frac{A}{E_{2}}) \\ P (A) = \frac{5}{11} \times \frac{3}{8} + \frac{6}{11} \times \frac{4}{8} \\ = \frac{15}{88} + \frac{24}{88} \\ = \frac{15 + 24}{88} = \frac{39}{88} \end{aligned}$

Probability exercise 30.6 question 2

Answer:
$\frac{19}{42}$
Hint:
We use the probability formula
Given:
Purse 1 = 2 silver ; 4 copper coins
Purse 2 = 4 silver ; 3 copper coins
Solution:
Let E₁ be the event of selecting purse 1 and then drawing a silver coin from it
Let E₂ be the event of selecting purse 2 and then drawing a silver coin from it
Let S be the event of drawing a silver coin
$\begin{aligned} P (E_{1}) = \frac{1}{2}; P (E_{2}) = \frac{1}{2} \\ P (\frac{S}{E_{1}}) = \frac{2}{6}; P (\frac{S}{E_{2}}) = \frac{4}{7} \end{aligned}$
Using total probability theorem
$\begin{aligned} required probability = P (E_{1}) \times P (\frac{S}{E_{1}}) + P (E_{2}) \times P (\frac{S}{E_{2}}) \\ = \frac{1}{2} \times \frac{2}{6} + \frac{1}{2} \times \frac{4}{7} \\ = \frac{1}{2} [\frac{2}{6} + \frac{4}{7}] \\ = \frac{1}{2} [\frac{1}{3} + \frac{4}{7}] \\ = \frac{1}{2} [\frac{7 + 12}{21}] = \frac{19}{42} \end{aligned}$

Probability exercise 30.6 question 3

Answer:
$\frac{29}{45}$
Hint:
To solve this problem, we use
$total probability = P (E_{1}) \times P (\frac{A}{E_{1}}) + P (E_{2}) \times P (\frac{A}{E_{2}})$
Given:
One bag contains 4 yellow and 5 red balls. Another bag contains 6 yellow and 3 red balls.
Solution:
E₁ = red bag from 1 to 2
E₂ = yellow bag from 1 to 2
A = ball is yellow
$\begin{aligned} P (E_{1}) = \frac{5}{9}; P (E_{2}) = \frac{4}{9} \end{aligned}$
Total balls in 2nd bag = 10
$\begin{aligned} P (\frac{A}{E_{1}}) = \frac{6}{10}; P (\frac{A}{E_{2}}) = \frac{7}{10} \end{aligned}$
Using total probability theorem
$\begin{aligned} required probability = P (A) = P (E_{1}) \times P (\frac{A}{E_{1}}) + P (E_{2}) \times P (\frac{A}{E_{2}}) \\ = \frac{5}{9} \times \frac{6}{10} + \frac{4}{9} \times \frac{7}{10} \\ = \frac{1}{3} + \frac{14}{45} \\ = \frac{15 + 14}{45} = \frac{29}{45} \end{aligned}$

Probability exercise 30.6 question 4

Answer:
$\frac{7}{15}$
Hint:
To solve this we use only white balls and then total probability formula
Given:
A bag contains 3 white and 2 black balls and another bag contains 2 white and 4 black balls.
Solution:
Bag 1 → 3W + 2B = 5 Bag 2 → 2W + 4B = 6
$\begin{aligned} P (Choosing a bag) = \frac{1}{2} \\ P (Bag 1) = P (Bag 2) = \frac{1}{2} \\ Bag 1 = White ball \\ P (W_{1}) = \frac{Number of white balls}{Total Number of balls} = \frac{3}{5} \\ Bag 2 = White ball \\ P (W_{1}) = \frac{Number of white balls}{Total Number of balls} = \frac{2}{6} = \frac{1}{3} \\ P (W) = \frac{1}{2} \times \frac{3}{5} = \frac{3}{10} \dots \dots probability of drawing white ball from Bag 1 \\ P (W) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} \dots \dots probability of drawing white ball from Bag 2 \\ Total probability = \frac{3}{10} + \frac{1}{6} = \frac{9 + 5}{30} = \frac{14}{30} = \frac{7}{15} \end{aligned}$

Probability exercise 30.6 question 5
Answer:
$\frac{118}{495}$
Hint:
To solve this we use total probability formula.
Given:
Bag I: 1 White,2 black and 3 red balls
Bag II: 2 White,1 black and 1 red balls
Bag III: 4 White,5 black and 3 red balls
Solution:
E₁=Selecting Bag 1
E₂=Selecting Bag 2
E₃=Selecting Bag 3
A = Drawing a white and a red ball
As one of the bag is selected randomly,
$\begin{aligned} P (E_{1}) = P (E_{2}) = P (E_{3}) = \frac{1}{3} \\ P (\frac{A}{E_{1}}) = \frac{C_{1} \times^{3} C_{1}}{^{6} C_{2}} = \frac{3}{15} \\ P (\frac{A}{E_{2}}) = \frac{^{2} C_{1} \times^{1} C_{1}}{^{4} C_{2}} = \frac{2}{6} \\ P (\frac{A}{E_{3}}) = \frac{^{4} C_{1} \times^{3} C_{1}}{^{12} C_{2}} = \frac{12}{66} \end{aligned}$
Using total probability theorem
$\begin{aligned} Required probability = P (E_{1}) \times P (\frac{A}{E_{1}}) + P (E_{2}) \times P (\frac{A}{E_{2}}) + P (E_{3}) \times P (\frac{A}{E_{3}}) \\ = \frac{1}{3} \times \frac{3}{15} + \frac{1}{3} \times \frac{2}{6} + \frac{1}{3} \times \frac{12}{66} \\ = \frac{1}{15} + \frac{1}{9} + \frac{2}{33} \\ = \frac{33 + 55 + 30}{495} = \frac{118}{495} \end{aligned}$

Probability exercise 30.6 question 6

Answer:
$\frac{193}{792}$
Hint:
To solve this first we find P(7 or 8) by head then P(7 & 8) by tails then required probability.
Given:
One coin – Head or Tail
Dice and 11 cards
Solution:
$Probability of an Head or Tail = (\frac{1}{2})$
Sample space of sum of the numbers of an unbiased dice when there will be head
$\begin{aligned} Sample space = {\begin{array}{c} 1 + 1; & 2 + 1 \dots & 6 + 1 \\ 1 + 2; & 2 + 2 \dots & 6 + 2 \\ 1 + 3; & 2 + 3 \dots & 6 + 1 \\ \dots \dots & \dots \dots & \dots \dots \\ 1 + 6; & 2 + 6 \dots & 6 + 6 \end{array} \\ n (S) = 36 \end{aligned}$
favourable outcome = 7 or 8 (Sum)
n favour = 11
$\begin{aligned} P (7 or 8) = \frac{11}{36} \end{aligned}$
Sample space of eleven cards when there will be tail,
Sample space = {2,3,4,5,6,7,8,9,10,11}
n(S)=11
favourable outcome = {7,8}, n favour = 2
$\begin{aligned} P (7 or 8) = \frac{2}{11} \\ P (7 or 8 after getting head) = \frac{1}{2} \times \frac{11}{36} = \frac{11}{72} \\ Required prob (7 or 8) = \frac{11}{72} + \frac{1}{11} \\ = \frac{121 + 72}{72 \times 11} \\ = \frac{193}{792} \end{aligned}$

Probability exercise 30.6 question 7

Answer:
0.016
Hint:
To solve this we use the total probability formula.
Given:
A=production 60%, 2% defective
B=production 40%, 1% defective
Solution:
$\begin{aligned} A = 60 = \frac{6}{10} \times \frac{2}{100} (that item is defective and coming from machine A ) \\ B = 40 = \frac{4}{10} \times \frac{1}{100} (that item is defective and coming from machine B ) \end{aligned}$
Using total probability theorem
$\begin{aligned} Required probability = \frac{6}{10} \times \frac{2}{100} + \frac{4}{10} \times \frac{1}{100} \\ = \frac{12 + 4}{1000} = \frac{16}{1000} = 0.016 \end{aligned}$

Probability exercise 30.6 question 8

Answer:
$\frac{83}{150}$
Hint:
To solve this we use total probability formula.
Given:
Bag A contains 8 white and 7 black balls & Bag B contains 5 white and 4 black balls
Solution:
E₁ = Black ball transfer from Bag A to Bag B
E₂ = White ball transfer from Bag A to Bag B
A = White ball drawn
$\begin{aligned} P (E_{1}) = \frac{7}{15}; P (E_{2}) = \frac{8}{15} \\ P (\frac{A}{E_{1}}) = \frac{5}{10}; P (\frac{A}{E_{2}}) = \frac{3}{5} \end{aligned}$
Using total probability theorem
$\begin{aligned} required probability = P (E_{1}) \times P (\frac{A}{E_{1}}) + P (E_{2}) \times P (\frac{A}{E_{2}}) \\ = \frac{7}{15} \times \frac{5}{10} + \frac{8}{15} \times \frac{3}{5} \\ = \frac{7}{30} + \frac{8}{25} \\ = \frac{83}{150} \end{aligned}$

Probability exercise 30.6 question 9

Answer:
$\frac{31}{72}$
Hint:
To solve this we find probability of white ball
Given:
Bag I contains 4 white and 5 black balls & Bag II contains 3 white and 4 black balls
Solution:
E₁ = Black ball transfer from first to second bag
E₁ = White ball transfer from first to second bag
A = White ball drawn
$\begin{aligned} P (E_{1}) = \frac{5}{9}; P (E_{2}) = \frac{4}{9} \\ P (\frac{A}{E_{1}}) = \frac{3}{8}; P (\frac{A}{E_{2}}) = \frac{4}{8} \end{aligned}$
Using total probability theorem
$\begin{aligned} Required probability = P (E_{1}) \times P (\frac{A}{E_{1}}) + P (E_{2}) \times P (\frac{A}{E_{2}}) \\ = \frac{5}{9} \times \frac{3}{8} + \frac{4}{9} \times \frac{4}{8} \\ = \frac{15}{72} + \frac{2}{9} \\ = \frac{15 + 16}{72} \\ = \frac{31}{72} \end{aligned}$

Probability exercise 30.6 question 10

Answer:
$\frac{29}{63}$
Hint:
To solve this we use total probability
Given:
Bag 1 → 4 white balls ; 5 black balls
Bag 2 → 6 white balls ; 7 black balls
Solution:
P ( white ball is drawn ) =
P ( white ball transfer ) x P( white ball is drawn ) + P ( black ball transfer ) x P( black ball is drawn )

\begin{aligned} = \frac{4}{9} \times \frac{7}{14} + \frac{5}{9} \times \frac{6}{14} \\ = \frac{28}{126} + \frac{30}{126} \\ = \frac{58}{126} = \frac{29}{63} \end{aligned}

Probability exercise 30.6 question 11

Answer:
$\frac{59}{130}$
Hint:
To solve this we use
$\begin{aligned} P (A) = P (E_{1}) \times P (\frac{A}{E_{1}}) + P (E_{2}) \times P (\frac{A}{E_{2}}) + P (E_{3}) \times P (\frac{A}{E_{3}}) \end{aligned}$
Given:
An urn contains 10 white and 3 black balls. Another urn contains 3 white and 5 black balls.
Solution:
E₁=2W …{2 white balls are transferred from I to II urn}
E₂=2B……{2 black balls are transferred from I to II urn}
E₃=1W,1B….{A white ball and a black ball are transferred from I to II urn}
A = White ball drawn
$\begin{aligned} P (E_{1}) = \frac{^{10} C_{2}}{^{13} C_{2}}; P (E_{2}) = \frac{^{3} C_{2}}{^{13} C_{2}}; P (E_{3}) = \frac{^{10} C_{1}^{3} C_{1}}{^{13} C_{2}} \\ A = white ball \\ P (\frac{A}{E_{1}}) = \frac{1}{2}; P (\frac{A}{E_{2}}) = \frac{3}{10}; P (\frac{A}{E_{3}}) = \frac{2}{5} \end{aligned}$
Using total probability theorem
$\begin{aligned} required probability = P (A) = P (E_{1}) \times P (\frac{A}{E_{1}}) + P (E_{2}) \times P (\frac{A}{E_{2}}) + P (E_{3}) \times P (\frac{A}{E_{3}}) \\ = \frac{10 \times 9}{13 \times 12} \times \frac{1}{2} + \frac{3 \times 2}{13 \times 12} \times \frac{3}{10} + \frac{30 \times 2}{13 \times 12} \times \frac{2}{5} \\ = \frac{15}{52} + \frac{3}{260} + \frac{2}{13} \\ = \frac{225 + 9 + 120}{780} = \frac{354}{780} = \frac{177}{390} \\ = \frac{59}{130} \end{aligned}$

Probability exercise 30.6 question 12

Answer:
$\frac{59}{105}$
Hint:
To solve this we make two case and then use total probability formula
Given:

Bag 1 – 6R + 8B

Bag 2 – 8R + 6B
Solution:
$\begin{aligned} Case 1: Red ball is transferred = \frac{6}{14} \times \frac{9}{15} \\ Case 2: Black ball is transferred = \frac{8}{14} \times \frac{8}{15} \\ P = Case 1 + Case 2 \\ P = \frac{6}{14} \times \frac{9}{15} + \frac{8}{14} \times \frac{8}{15} \\ = \frac{54 + 64}{14 \times 15} = \frac{118}{14 \times 15} = \frac{118}{210} \\ = \frac{59}{105} \end{aligned}$

Probability exercise 30.6 question 13

Answer:
$\frac{17}{400}$
Hint:
To solve this we find the (B₁, B₂ and B₃) then use total probability
Given:
E₁: 50% Production, E₂: 25% Production, E₃: 25% Production
Solution:
B₁ is the event when E₁ produce bulbs
B₂ is the event when E₂ produce bulbs
B₃ is the event when E₃ produce bulbs
E event when a defective bulbs is selected
$\begin{aligned} P (B_{1}) = \frac{1}{2}; P (B_{2}) = \frac{1}{4}; P (B_{3}) = \frac{1}{4} \\ P (\frac{E}{B_{1}}) = \frac{4}{100} = \frac{1}{25} \\ P (\frac{E}{B_{2}}) = \frac{4}{100} = \frac{1}{25} \\ P (\frac{E}{B_{3}}) = \frac{5}{100} = \frac{1}{20} \end{aligned}$
Using total probability theorem,
$\begin{aligned} P (E) = P (B_{1}) \times P (\frac{E}{B_{1}}) + P (B_{2}) \times P (\frac{E}{B_{2}}) + P (B_{3}) \times P (\frac{E}{B_{3}}) \\ P (E) = \frac{1}{2} \times \frac{1}{25} + \frac{1}{4} \times \frac{1}{25} + \frac{1}{4} \times \frac{1}{20} \\ P (E) = \frac{17}{400} \end{aligned}$

RD Sharma class 12 solutions Probability 30.6 can be a holy grail for students in their exam preparations. Chapter 30 of the NCERT will teach students about Probability. Exercise 30.6 has 13 questions that cover the concepts of conditional probability, and the Bayes theorem.

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