RD Sharma Class 12 Exercise 30.5 Probability Solutions Maths-Download PDF Online

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RD Sharma Class 12 Exercise 30.5 Probability Solutions Maths-Download PDF Online

RD Sharma Class 12 Exercise 30.5 Probability Solutions Maths-Download PDF Online

Satyajeet KumarUpdated on 25 Jan 2022, 04:02 PM IST

RD Sharma Solutions are the enthusiastically suggested reference material by the majority of the CBSE schools. Without even a trace of an instructor, this book will direct the students the correct way by explaining their doubts and giving detailed replies. The Class 12 RD Sharma Chapter 30 Exercise 30.5 is the best booklet for any student to prepare their exam with. RD Sharma Solutions Students can try many mock tests. Working on self-assessments using the RD Sharma CLass 12th Exercise 30.5 increases the capability and confidence of the students to face the exams.
The RD Sharma class 12 solutions of Probability exercise 30.5, is utilized by thousands of students and educators for useful information on maths. The RD Sharma class 12th exercise 30.5 comprises of 56 inquiries covering every one of the fundamental ideas identified with requests identified with cards, flip a coin, two balls an and b, discovering probability of at any rate, probably, thrown a dice, and so forth and this inquiry depended on Binomial, Autonomous occasions, Association, and crossing point, with substitution and without substitution.

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RD Sharma Class 12 Solutions Chapter30 Probability - Other Exercise
Probability Excercise: 30.5
RD Sharma Chapter-wise Solutions

RD Sharma Class 12 Solutions Chapter30 Probability - Other Exercise

Probability Excercise: 30.5

Probaility Exercise 30.5 Question 1

Answer:$\frac{14}{27}$
Hint: You must know the rules of finding probability
Given: Bag 1 – 3 white & 6 black balls
Bag 2 – 5 black & 4 white balls
Solution:
$Bag 1=\left ( 3W+6B \right )=9balls$
$Bag 2=\left ( 5B+4W \right )=9balls$
$P\left (balls\: of\: same\: colour\: are\: drawn \right )=P\left ( both\: black \right )+P\left ( both\: white \right )$
$\begin{aligned} &=\left(\frac{6}{9} \times \frac{5}{9}\right)+\left(\frac{3}{9} \times \frac{4}{9}\right) \\ &=\frac{30}{81}+\frac{12}{81} \\ &=\frac{42}{81} \\ &P=\frac{14}{27} \end{aligned}$

Probability Exercise 30.5 Question 2

Answer:$\frac{21}{40}$
Hint: You must know the rules of finding probability
Given: Bag 1 $\rightarrow$ 3 Red and 5 Black balls
Bag 2 $\rightarrow$ 6 Red and 4 Black balls
Solution:
Bag 1$\rightarrow$(3R+5B) = 8 balls
Bag 2 $\rightarrow$(6R+4B) =10 balls
$\begin{aligned} &\mathrm{P}(\text { one is red and one is black })=\mathrm{P}(\text { red from bag } 1 \text { and black from } 2 \text { bag) } \\ \end{aligned}$
$\begin{aligned} &+\mathrm{P}(\text { red from bag } 2 \text { and black from } 1 \text { bag }) \\ \end{aligned}$
$\begin{aligned} &=\left(\frac{3}{8} \times \frac{4}{10}\right)+\left(\frac{5}{8} \times \frac{6}{10}\right) \\ &=\frac{12}{80}+\frac{30}{80} \\ &=\frac{42}{80} \\ \end{aligned}$
$\begin{aligned} &P=\frac{21}{40} \end{aligned}$

Probability Exercise 30.5 Question 3 (i)

Answer:$\frac{16}{81}$
Hint: You must know the rules of finding probability
Given: Box contains 10 black and 8 red balls
Solution: Box = (10 B and 8 R) balls
Total no. of balls = ( 10 +8) =18 balls
$P\left ( both\: red\: balls \right )$
$=\left ( \frac{8}{18}\times \frac{8}{18} \right )$
$=\frac{64}{18\times 18}$
$P=\frac{16}{81}$

Probability Exercise 30.5 Question 3 (iii)

Answer:$\frac{40}{81}$

Hint: You must know the rules of finding probability

Given:Box contains 10 black and 8 red balls

Solution:Box = (10B and 8R) balls

Total no. of balls =10 + 8 = 18 balls

P (one is black and other is red) = P(first red ball and second black)+P(first black ball and second red )

$=\left ( \frac{8}{18}\times \frac{10}{18} \right )+\left ( \frac{10}{18}\times \frac{8}{18} \right )$

$=\frac{80}{324}+\frac{80}{324}$

$=\frac{160}{324}$

$=\frac{40}{81}$

Probability Exercise 30.5 Question 4

Answer:$\frac{32}{221}$
Hint: You must know the rules of finding probability .
Given: Two cards are drawn from 52 cards without replacement
Solution: P(exactly one ace)=P(first card ace)+P(second not ace)
$\begin{aligned} =&\left(\frac{4}{52} \times \frac{48}{51}\right)+\left(\frac{48}{52} \times \frac{4}{51}\right) \\ =& \frac{192+192}{52 \times 51} \\ =& \frac{384}{52 \times 51} \\ =& \frac{32}{221} \end{aligned}$

Probability Exercise 30.5 Question 5

Answer: $35%$

Hint: You must know the rules of finding probability

Given: A speaks $75%$ of truth

B speaks 80% of truth

Solution:P(both narrating different incidents)=P(A speaks truth and B lies)+P(A lies and B speaks truth)

$\begin{aligned} &=P(A \cap \bar{B})+P(\bar{A} \cap B) \\ &{[\because P(A \cap B)=P(A) P(B)]} \\ &=P(A) P(\bar{B})+P(\bar{A}) P(B)] \\ &{[P(\bar{B})=1-P(B)]} \\ \end{aligned}$

$\begin{aligned} &=0.75(1-0.80)+(1-0.75)(0.80) \\ &=0.75(0.20)+(0.25)(0.80) \\ &=0.15+0.2 \\ &=0.35 \\ &=35 \% \end{aligned}$

Probability Exercise 30.5 Question 6 (i)

Answer:$\frac{1}{15}$
Hint: You must know the rules of finding probability
Given: Probability of Kamal’s selection$=\frac{1}{3}$
Probability of Monika’s selection$=\frac{1}{5}$
Solution: P( Kamal selected)$=\frac{1}{3}=P\left ( A \right )$
P( Monika selected)$=\frac{1}{5}=P\left ( B \right )$
$P\left ( both\: get\: selected \right )=P\left ( A \right )\times P\left ( B \right )$
$=\frac{1}{3}\times \frac{1}{5}$
$=\frac{1}{15}$

Probability Exercise 30.5 Question 6 (ii)

Answer:$\frac{8}{15}$
Hint: You must know the rules of finding probability
Given:Probability of Kamal’s selection$=\frac{1}{3}$
Probability of Monika’s selection$=\frac{1}{5}$
Solution:P( Kamal selected)$=\frac{1}{3}=P\left ( A \right )$
P( Monika selected)$=\frac{1}{5}=P\left ( B \right )$
$\begin{aligned} &\mathrm{P}(\text { none of them get selected })=P(\bar{A}) \times P(\bar{B}) \\ &=[1-P(A)][1-P(B)] \\ &=\left(1-\frac{1}{3}\right) \times\left(1-\frac{1}{5}\right) \\ &=\frac{2}{3} \times \frac{4}{5} \\ &=\frac{8}{15} \end{aligned}$

Probability Exercise 30.5 Question 6 (iii)

Answer:$\frac{7}{15}$
Hint: You must know the rules of finding probability
Given:Probability of Kamal’s selection$=\frac{1}{3}$
Probability of Monika’s selection$=\frac{1}{5}$
Solution:P( Kamal selected)$=\frac{1}{3}=P\left ( A \right )$
P( Monika selected)$=\frac{1}{5}=P\left ( B \right )$
$\begin{aligned} &\mathrm{P}(\text { atleast one get selected })=P(A \cup B) \\\\ &P(A \cup B)=P(A)+P(B)-P(A \cap B) \\\\ &=P(A)+P(B)-P(A) P(B) \\ \end{aligned}$
$\begin{aligned} &=\frac{1}{3}+\frac{1}{5}-\left(\frac{1}{3} \times \frac{1}{5}\right) \\ &=\frac{1}{3}+\frac{1}{5}-\left(\frac{1}{15}\right) \\ &=\frac{5+3-1}{15} \\ &=\frac{7}{15} \end{aligned}$

Probability Exercise 30.5 Question 6 (iv)

Answer:$\frac{2}{5}$
Hint: You must know the rules of finding probability
Given:Probability of Kamal’s selection$=\frac{1}{3}$
Probability of Monika’s selection$=\frac{1}{5}$
Solution:P( Kamal selected)$=\frac{1}{3}=P\left ( A \right )$
P( Monika selected)$=\frac{1}{5}=P\left ( B \right )$
$\begin{aligned} &\mathrm{P} \text { (one of them gets selected) }=P(\bar{A}) P(B)+P(\bar{B}) P(A) \\\\ &{\left[\begin{array}{l} P(\bar{A}) \text { means A is not selected } \\\\ P(\bar{B}) \text { means B is not selected } \end{array}\right]} \\\\ &=P(B)[1-P(A)]+P(A)[1-P(B)] \\ \end{aligned}$
$\begin{aligned} &=\frac{1}{5}\left(1-\frac{1}{3}\right)+\frac{1}{3}\left(1-\frac{1}{5}\right) \\ &=\frac{2}{15}+\frac{4}{15} \\ &=\frac{6}{15} \\ &=\frac{2}{5} \end{aligned}$

Probability Exercise 30.5 Question 7

Answer:$\frac{5}{22}$
Hint: You must know the rules of finding probability.
Balls are drawn without replacement.
Given: Box contains (3White + 4Red + 5Black) balls
Solution: Box=(3W+4R+5B)
Total no. of balls = 3+ 4+ 5=12 balls
P(one white and one black)=P(first white, second black)+P(first black, second white)
$\begin{aligned} &=\frac{3}{12} \times \frac{5}{11}+\frac{5}{12} \times \frac{3}{11} \\ &=\frac{15}{132}+\frac{15}{132} \\ &=\frac{30}{132} \\ &=\frac{5}{22} \end{aligned}$

Probability Exercise 30.5 Question 8

Answer:$\frac{5}{13}$
Hint: You must know the rules of finding probability.
Given :Bag contains 8 red and 6 green balls.
The balls are drawn without replacement.
Solution: Bag = (8R + 6G) =14 balls
P(atleast 2 balls are green)= 1-P(at most one ball is green)
=1-[P(first green, second and third red)+ P(first red , second green , third red ) +P(first and second red , third green) +P( all red)]
$\begin{aligned} &=1-\left[\frac{6}{14} \times \frac{8}{13} \times \frac{7}{12}+\frac{8}{14} \times \frac{6}{13} \times \frac{7}{12}+\frac{8}{14} \times \frac{7}{13} \times \frac{6}{12}+\frac{8}{14} \times \frac{7}{13} \times \frac{6}{12}\right] \\\\ &=1-\left[\frac{336}{2184}+\frac{336}{2184}+\frac{336}{2184}+\frac{336}{2184}\right] \\\\ &=1-\frac{1344}{2184} \\\\ &=\frac{2184-1344}{2184}=\frac{840}{2184} \\\\ &=\frac{5}{13} \end{aligned}$

Probability Exercise 30.5 Question 9

Answer:$\frac{1}{2}$
Hint: You must know the rules of finding probability
Given:
$Probability \: \: of \: \: Arun's \: \: selection=\frac{1}{4}=P\left ( A \right )$
$Probability \: \: of \: \: Tarun's \: \: rejection=\frac{2}{3}=P\left ( \bar{B} \right )$
Solution:
$\begin{aligned} &P(\text { Arun's selected })=\frac{1}{4} \\ &P(\text { Tarun's rejected })=\frac{2}{3} \\ &P(\text { Tarun's selected })=\frac{1}{1}-\frac{2}{3}=\frac{1}{3} \end{aligned}$
$\begin{aligned} &\mathrm{P}(\text { atleast one of them gets selected })=P(A \cup B) \\\\ &P(A \cup B)=P(A)+P(B)-P(A \cap B) \\\\ &=P(A)+P(B)-P(A) P(B) \\ \end{aligned}$
$\begin{aligned} &=\frac{1}{4}+\frac{1}{3}-\frac{1}{4} \times \frac{1}{3} \\ &=\frac{1}{4}+\frac{1}{3}-\frac{1}{12} \\ &=\frac{3+4-1}{12}=\frac{6}{12} \\ &=\frac{1}{2} \end{aligned}$

Probability Exercise 30.5 Question 10

Answer:$\frac{1}{3}$
Hint: You must know the rules of finding probability and G.P series
Given: A and B toss a coin alternately till one of them get a heads and wins thegame
Solution: If A starts the game, then B will win only at even positions
P(winning the game) = P(head at 2^nd turn + 4^th turn + …….)
$\begin{aligned} &\Rightarrow\left(\frac{1}{2} \times \frac{1}{2}\right)+\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)+\ldots \ldots \ldots . \\\\ &=\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{4}+\left(\frac{1}{2}\right)^{6}+\ldots \ldots . . \\\\ &\Rightarrow \frac{1}{4}\left[1+\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{4}+\ldots .\right] \\\\ &\Rightarrow \frac{1}{4}\left[\frac{1}{1-\frac{1}{4}}\right] \quad\left[\text { For G.P: } 1+a+a^{2}+a^{3}+\ldots=\frac{1}{1-a}\right] \\\\ &\Rightarrow \frac{1}{4} \times \frac{4}{3} \\\\ &\Rightarrow \frac{1}{3} \end{aligned}$

Probability Exercise 30.5 Question 11

Answer:$\frac{26}{51}$
Hint: You must know the rules of finding probability functions.
Given: Two cards are drawn from 52 cards one after another without replacement.
Solution:
P(one red and other black)= P(first red and second black) + P(first black and second red)
$=\frac{26}{52} \times \frac{26}{51}+\frac{26}{52} \times \frac{26}{51} \ldots . with\: \: out \: \: replacement$
$\begin{aligned} &=\frac{13}{51}+\frac{13}{51} \\ &=\frac{26}{51} \end{aligned}$

Probability Exercise 30.5 Question 12

Answer: $\frac{4}{45}$
Hint: You must know the rules of finding probability functions.
Given: Tickets are number from 1 to 10
Two tickets are drawn one after the other at random.
Solution: We know 5 and 10 are multiples of 5 while 4 and 8 are multiples of 4
$\begin{aligned} &P(\text { Multiple of } 5)=\frac{2}{10}=\frac{1}{5} \\ &P(\text { Multiple of } 4)=\frac{2}{10}=\frac{1}{5} \end{aligned}$
$P(Multiple\: of \: 5\: and \: multiply \: of \: 4)=P( multiple\: of \: 5 \: on \: first and \: multiple \: of \: 4 \: on \: second)+$
$P(multiple of 4 on first and multiple of 5 on second)$
$\begin{aligned} &=\left(\frac{2}{10} \times \frac{2}{9}\right)+\left(\frac{2}{10} \times \frac{2}{9}\right) \\ \end{aligned}$ $\left [ without\: replacement \right ]$

$\begin{aligned} &=\frac{4}{90}+\frac{4}{90} \\ &=\frac{8}{90} \\ &=\frac{4}{45} \end{aligned}$

Probaility Exercise 30.5 Question 13

Answer:$44%$
Hint: You must know the rules of finding probability functions.
Given: Husband tells a lie in $30%$ cases,
Wife tells a lie in $35%$ cases.
Solution:
$\begin{aligned} &P(\text { husband lies })=\frac{30}{100}=0.3 \\ &P(\text { wife lies })=\frac{35}{100}=0.35 \end{aligned}$
$P(both\: will \: contradict \: each\: other\: on \: the\: same\: fact) =$
$P(husband \: lies,\: wife\: not) + P(wife \: lies, husband\: not)$
$\begin{aligned} &P(H) P(\bar{W})+P(W) P(\bar{H}) \\ \end{aligned}$
$\begin{aligned} &=0.3 \times(1-0.35)+(1-0.3) \times 0.35 \\ &=(0.3 \times 0.65)+(0.7 \times 0.35) \\ &=0.195+0.245 \\ &=0.44 \\ &\therefore \text { Both contradict }=44 \% \end{aligned}$

Probability Exercise 30.5 Question 14 (i)

Answer:$\frac{1}{35}$
Hint: You must know the rules of finding probability.
Given:
$Probability of husband will be selected =\frac{1}{7}=P(A)$
$Probability of wife will be selected =\frac{1}{5}=P(B)$
Solution:
$\begin{aligned} &P(\text { husband selected })=\frac{1}{7} \\ &P(\text { wife selected })=\frac{1}{5} \\ &P(\text { both will be selected })=P(A \cap B) \\ &=P(A) P(B) \\ &=\frac{1}{7} \times \frac{1}{5} \\ &=\frac{1}{35} \end{aligned}$

Probability Exercise 30.5 Question 14 (ii)

Answer:$\frac{2}{7}$
Hint: You must know the rules of finding probability.
Given:
$Probability of husband's will selected =\frac{1}{7}=P(A)$
$Probability of wife will selected =\frac{1}{5}=P(B)$
Solution:
$\begin{aligned} &P(\text { husband selected })=\frac{1}{7} \\ &P(\text { wife selected })=\frac{1}{5} \end{aligned}$
$\begin{aligned} &P(\text { one of them will be selected })=P(A) P(\bar{B})+P(\bar{A}) P(B) \\ \end{aligned}$
$\begin{aligned} &=\frac{1}{7}\left(1-\frac{1}{5}\right)+\frac{1}{5}\left(1-\frac{1}{7}\right) \\ &=\frac{1}{7}\left(\frac{4}{5}\right) \times \frac{1}{5}\left(\frac{6}{7}\right) \\ &=\frac{4}{35}+\frac{6}{35} \\ &=\frac{10}{35} \\ &=\frac{2}{7} \end{aligned}$

Probability Exercise 30.5 Question 14 (iii)

Answer:

Answer:$\frac{24}{35}$
Hint: You must know the rules of finding probability.
Given:
$Probability of husband's will selected =\frac{1}{7}=P(A)$
$Probability of wife will selected =\frac{1}{5}=P(B)$
Solution:
$\begin{aligned} &P(\text { husband selected })=\frac{1}{7} \\\\ &P(\text { wife selected })=\frac{1}{5} \\\\ &P(\text { none of them will be selected })=P(\overline{A \cap B}) \\\\ &=P(\bar{A}) \times P(\bar{B}) \\\\ &=\left(1-\frac{1}{7}\right) \times\left(1-\frac{1}{5}\right) \\\\ &=\left(\frac{6}{7}\right)\left(\frac{4}{5}\right) \end{aligned}$
$=\frac{24}{35}$

Probability Exercise 30.5 Question 15.

Answer:$\frac{23}{364}$
Hint: You must know the rules of finding probability.
Given: Bag contains 7 white, 5 black, 4 red balls
Four balls are drawn without replacement.
Solution
We have
White=7, Black=5, Red=4
Four balls are drawn without replacement
P(at least 3 black balls are black)
=P(3 black balls and one not black or 4 black balls)
=P(3 black and one not black)+P(4 black balls)
$\begin{aligned} &n C_{r}=\frac{n !}{r !(n-r) !} \\ &=\frac{5 C_{3} \times 11 C_{1}}{16 C_{4}}+\frac{5 C_{4}}{16 C_{4}} \\ &=\frac{51}{31(5-3)} \times \frac{111}{11(11-1)}+\frac{51}{41(5-4)} \\ \end{aligned}$
$\begin{aligned} &=\frac{16 !}{4 !(16-4) !} \\\\ &=\frac{5 !}{3 ! 2 !} \times \frac{11 !}{10 !}+\frac{5 !}{4 ! 1 !} \\\\ &=\frac{16 !}{4 ! 2 !} \\\\ &=\frac{\frac{5.4}{2} \times 11+5}{\frac{16.15 .14 .13}{4.3 .2}} \end{aligned}$
$=\frac{\left ( 110+5 \right )}{1820}$
$=\frac{115}{1820}$
$=\frac{23}{364}$

Probability Exercise 30.5 Question 16

Answer:$\frac{107}{120}$
Hint: You must know the rules of finding probability.
Given: A speaks truth three times out of four
B speaks truth four times out of five
C speaks truth five times out of six
Solution:$P\left ( A\: Speaks\: truth \right )=\frac{3}{4}$
$P\left ( B\: Speaks\: truth \right )=\frac{4}{5}$
$P\left ( C\: Speaks\: truth \right )=\frac{5}{6}$
$\begin{aligned} &P \text { (majority speaks truth) }=P \text { (two speak truth) }+P(\text { all speak truth }) \\ \end{aligned}$
$\begin{aligned} &=P(A) \times P(B)[1-P(C)]+P(A) \times P(C)[1-P(B)] \\ \end{aligned}$$\begin{aligned} &+P(C) \times P(B)[1-P(A)]+P(A) \times P(B) P(C) \\ \end{aligned}$$\begin{aligned} &+P(C) \times P(B)[1-P(A)]+P(A) \times P(B) P(C) \\ \end{aligned}$
$\begin{aligned} &=\frac{3}{4} \times \frac{4}{5}\left(1-\frac{5}{6}\right)+\frac{3}{4} \times \frac{5}{6}\left(1-\frac{4}{5}\right)+\frac{4}{5} \times \frac{5}{6}\left(1-\frac{3}{4}\right)+\frac{3}{4} \times \frac{4}{5} \times \frac{5}{6} \\ &=\frac{12}{120}+\frac{15}{120}+\frac{20}{120}+\frac{60}{120} \\ &=\frac{107}{120} \end{aligned}$

Probability Exercise 30.5 Question 17 (i)

Answer:$\frac{1}{4}$
Hint: You must know the rules of finding probability.
Given:
$Bag\left ( 1 \right )\rightarrow \left ( 4W+2B \right )balls$
$Bag\left ( 2 \right )\rightarrow \left ( 3W+5B \right )balls$
Solution:

Total no. of balls = 6 balls in bag (1)

Total no. of balls =8 balls in bag (2)

$Bag\left ( 1 \right )\rightarrow \left ( 4W+2B \right )balls$

$Bag\left ( 2 \right )\rightarrow \left ( 3W+5B \right )balls$

$P\left ( both \: are\: white \right )=\frac{4}{6}\times \frac{3}{8}=\frac{1}{4}$

Probability Exercise 30.5 Question 17 (ii)

Answer:$\frac{5}{24}$
Hint: You must know the rules of finding probability functions.
Given:
$Bag\left ( 1 \right )\rightarrow \left ( 4W+2B \right )balls$
$Bag\left ( 2 \right )\rightarrow \left ( 3W+5B \right )balls$
Solution:

Total no. of balls = 6 balls in bag (1)

Total no. of balls =8 balls in bag (2)

$Bag\left ( 1 \right )\rightarrow \left ( 4W+2B \right )balls$

$Bag\left ( 2 \right )\rightarrow \left ( 3W+5B \right )balls$

$P\left ( both\: are\: black \right )=\frac{2}{6}\times \frac{5}{8}=\frac{5}{24}$

Probability Exercise 30.5 Question 17 (iii)

Answer:$\frac{13}{24}$
Hint: You must know the rules of finding probability functions.
Given:
$Bag\left (1 \right )\rightarrow \left ( 4W+2B \right )balls$
$Bag\left (2 \right )\rightarrow \left ( 3W+5B \right )balls$

Solution:

Total no. of balls = 6 balls in bag (1)

Total no. of balls =8 balls in bag (2)

$Bag\left (1 \right )\rightarrow \left ( 4W+2B \right )balls$

$Bag\left (2 \right )\rightarrow \left ( 3W+5B \right )balls$

$\mathrm{P}( one\: is \: white \: and \: one \: is \: black) =$

$\mathrm{P}(white from bag 1 and black from 2 bag) +\mathrm{P} (black from bag 1 and white from 1 bag)$

$\begin{aligned} &=\left(\frac{4}{6} \times \frac{5}{8}\right)+\left(\frac{2}{8} \times \frac{3}{6}\right) \\ &=\frac{20}{48}+\frac{6}{48} \\ &=\frac{26}{48} \\ &=\frac{13}{24} \end{aligned}$

Probability Exercise 30.5 Question 18

Answer:$\frac{67}{256}$
Hint: You must know the rules of finding probability functions.
Given: Bag contains = (4W+5R+7B) balls
Solution
No. of white balls = 4

No. of black balls = 7
No. of red balls = 5

Total balls = 16

No. of ways in which 4 balls can be drawn from 16 balls$=16C_{4}$

Let A = getting at least two white balls = getting 2, 3, 4 white balls

No. of ways of choosing 2 white balls $=4C_{2}\times 12C_{2}$

No. of ways of choosing 3 white balls $=4C_{3}\times 12C_{1}$

No. of ways of choosing 4 white balls $=4C_{4}\times 12C_{0}$

$\begin{aligned} &P(A)=\frac{4 C_{2} \times 12 C_{2}+4 C_{3} \times 12 C_{1}+4 C_{4} \times 12 C_{0}}{16 C_{4}} \\ &=\frac{4 !}{2 ! 2 !} \times \frac{12 !}{2 ! 10 !}+\frac{4 !}{3 !} \times \frac{12 !}{11 !}+\frac{4 !}{4 ! 0 !} \times \frac{12 !}{12 !} \\ &=\frac{16 !}{4 ! 2 !} \\ &=\frac{67}{256} \end{aligned}$

Probability Exercise 30.5 Question 20 (i)

Answer:$\frac{43}{90}$
Hint: You must know the rules of finding probability functions.
Given:
$Bag \left ( A \right )\rightarrow \left ( 4R+5B \right )balls$
$Bag \left ( B \right )\rightarrow \left ( 3R+7B \right )balls$

Solution:

Total no. of balls = 9 balls in bag A

Total no. of balls =10 balls in bag B

P(balls of different colors) =

P (red from bag A and black from bag B) + P (red from bag B + black from bag A)

$=\left ( \frac{4}{9}\times \frac{7}{10} \right )+\left ( \frac{3}{10}\times \frac{5}{9} \right )$

$=\frac{28}{90}+\frac{15}{90}$

$=\frac{43}{90}$

Probability Exercise 30.5 Question 20 (ii)

Answer: $\frac{47}{90}$
Hint: You must know the rules of finding probability.
Given:
$Bag\left ( A \right )\rightarrow \left ( 4R+5B \right )balls$
$Bag\left ( B \right )\rightarrow \left ( 3R+7B \right )balls$
Solution:
Total no. of balls = 9 balls in bag A
Total no. of balls =10 balls in bag B
P (balls of same colors) = P (both red) + P (both black )
$=\left ( \frac{4}{9}\times \frac{3}{10} \right )+\left ( \frac{7}{10}\times \frac{5}{9} \right )$
$=\frac{12}{90}+\frac{35}{90}$
$=\frac{47}{90}$

Probability Exercise 30.5 Question 22 (i)

Answer: $\frac{8}{81}$
Hint: You must know the rules of finding probability functions.
Given:
$P\left ( A\: passing\: the\: exam \right )=\frac{2}{9}$
$P\left ( B\: passing\: the\: exam \right )=\frac{5}{9}$
Solution:
P(only A passing the examination) = P (A passes) P (B fails)
$\begin{aligned} &=\mathrm{P}(\mathrm{A} \cap \overline{B)} \\ &=\frac{2}{9}\left(1-\frac{5}{9}\right) \\ &=\frac{2}{9}\left(\frac{4}{9}\right) \\ &=\frac{8}{81} \end{aligned}$

Probability Exercise 30.5 Question 22 (ii)

Answer: $\frac{43}{81}$
Hint: You must know the rules of finding probability .
Given:
$P\left ( A\: passing\: the\: exam \right )=\frac{2}{9}$
$P\left ( B\: passing\: the\: exam \right )=\frac{5}{9}$

Solution:

P(only one of them passing the examination) = P (A passes ,B fails) + P (B passes , A fails)

$\begin{aligned} &=\frac{2}{9}\left(1-\frac{5}{9}\right)+\frac{5}{9}\left(1-\frac{2}{9}\right) \\ &=\frac{2}{9}\left(\frac{4}{9}\right)+\frac{5}{9}\left(\frac{7}{9}\right) \\ &=\frac{8}{81}+\frac{35}{81} \\ &=\frac{43}{81} \end{aligned}$

Probability Exercise 30.5 Question 23

Answer:$\frac{17}{42}$
Hint: You must know the rules of finding probability functions.
Given: Urn A = (4R+3B) = 7 balls
Urn B = (5R+4B) = 9 balls
Urn C = (4R+4B) = 8 balls
Solution
Urn A contain 4 red $\left ( R_{1} \right )$, 3 black $\left ( B_{1} \right )$ balls
Urn B contain 5 red $\left ( R_{2} \right )$, 4 black $\left ( B_{2} \right )$ balls
Urn B contain 4 red $\left ( R_{3} \right )$, 4 black $\left ( B_{3} \right )$ balls
P(3 balls drawn contain 2 red and a black ball)
$\begin{aligned} &=P\left[\left(R_{1} \cap R_{2} \cap B_{3}\right) \cup\left(R_{1} \cap B_{2} \cap R_{3}\right) \cup\left(B_{1} \cap R_{2} \cap R_{3}\right)\right] \\\\ &=P\left(R_{1} \cap R_{2} \cap B_{3}\right)+P\left(R_{1} \cap B_{2} \cap R_{3}\right)+P\left(B_{1} \cap R_{2} \cap R_{3}\right) \\\\ &=P\left(R_{1}\right) \times P\left(R_{2}\right) \times P\left(B_{3}\right)+P\left(R_{1}\right) \times P\left(B_{2}\right) \times P\left(R_{3}\right)+P\left(B_{1}\right) \times P\left(R_{2}\right) \times P\left(R_{3}\right) \\\\ &=\frac{4}{7} \times \frac{5}{9} \times \frac{4}{8}+\frac{4}{7} \times \frac{4}{9} \times \frac{4}{8}+\frac{3}{7} \times \frac{5}{9} \times \frac{4}{8} \\\\ &=\frac{80+64+60}{504} \\\\ &=\frac{204}{504} \\\\ &=\frac{17}{42} \end{aligned}$

Probability Exercise 30.5 Question 24 (i)

Answer:$0.03$
Hint: Probability required $=P\left ( A\cap B\cap C \right )$
Given:
$P\left ( A\: grade\: in\: Maths \right )=0.2=P\left ( A \right )$
$P\left ( A\: grade\: in\: Physics \right )=0.3=P\left ( B \right )$
$P\left ( A\: grade\: in\: Chemistry \right )=0.5=P\left ( C \right )$
Solution:$P\left ( A\: grade\: in\: all\: Subjects \right )=P\left ( A \right )\times P\left ( B \right )\times P\left ( C \right )$
$=0.2\times 0.3\times 0.5$
$=0.03$

Probability Exercise 30.5 Question 24 (ii)

Answer:$0.28$
Hint:$P\left ( \bar{A} \right )=1-P\left ( A \right )$
Given:
$P\left ( {A} \: grade\: in\: maths\right )=0.2=P\left ( A \right )$
$P\left ( {A} \: grade\: in\: Physics\right )=0.3=P\left ( B \right )$
$P\left ( {A} \: grade\: in\: Chemistry\right )=0.5=P\left ( C \right )$

Solution: P(grade A in no subject)

$\begin{aligned} &=P(\bar{A}) \times P(\bar{B}) \times P(\bar{C}) \\ &=(1-0.2) \times(1-0.3) \times(1-0.5) \\ &=0.8 \times 0.7 \times 0.5 \\ &=0.28 \end{aligned}$

Probability Exercise 30.5 Question 24 (iii)

Answer: $0.22$
Hint: You must know the rules of finding probability functions.
Given:
$\begin{aligned} &P(\mathrm{~A} \text { grade in Maths })=0.2=P(A) \\ &P(\mathrm{~A} \text { grade in Physics })=0.3=P(B) \\ &P(\mathrm{~A} \text { grade in Chemistry })=0.5=P(C) \end{aligned}$

Solution:

P(grade A in two subjects) =

P(not grade A in Math )+P(not grade A in Physics)+P(not grade A in Chemistry)

$\begin{aligned} &=P(\bar{A}) P(B) P(C)+P(A) P(\bar{B}) P(C)+P(A) P(B) P(\bar{C}) \\\\ &=[(1-0.2) \times 0.3 \times 0.5] \times[0.2 \times(1-0.3) \times 0.5] \times[0.2 \times 0.3 \times(1-0.5)] \\\\ &=[0.8 \times 0.3 \times 0.5] \times[0.2 \times 0.7 \times 0.5] \times[0.2 \times 0.3 \times 0.5] \\\\ &=0.12 \times 0.07 \times 0.03 \\\\ &=0.22 \end{aligned}$

Probability Exercise 30.5 Question 25

Answer:$\frac{9}{8}$
Hint: You must know the rules of finding probability functions.
Given: A and B take turns in throwing two dice, the first to throw 9 being awarded the prize
Solution: Total number of events = 36
P(getting 9) $=\frac{4}{36}=\frac{1}{9}$

P(A winning) = P(getting 9 in first throw) + P(getting 9 in third row)+….

$\begin{aligned} &=\frac{1}{9}+\left(1-\frac{1}{9}\right)\left(1-\frac{1}{9}\right) \times \frac{1}{9}+\ldots . . \\\\ &=\frac{1}{9}\left[1+\frac{64}{81}+\left(\frac{64}{81}\right)^{2}+\ldots\right] \\\\ &=\frac{1}{9}\left[\frac{1}{1-\frac{64}{81}}\right] \quad\left[1+a+a^{2}+a^{3}+\ldots=\frac{1}{1-a}\right] \\\\ &=\frac{1}{9}\left[\frac{81}{17}\right] \\\\ &=\frac{9}{17} \end{aligned}$

P(B winning)=P(getting 9 in second throw)+P(getting 9 in fourth throw)+…

$\begin{aligned} &=\left(1-\frac{1}{9}\right) \frac{1}{9}+\left(1-\frac{1}{9}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{9}\right) \times \frac{1}{9}+\ldots . . \\\\ &=\frac{8}{81}\left[1+\frac{64}{81}+\left(\frac{64}{81}\right)^{2}+\ldots .\right] \\\\ &=\frac{8}{81}\left[\frac{1}{1-\frac{64}{81}}\right] \quad\left[1+a+a^{2}+a^{3}+\ldots=\frac{1}{1-a}\right] \\\\ &=\frac{8}{81}\left[\frac{81}{17}\right] \\\\ &=\frac{8}{17} \end{aligned}$

Winning ratio of A to B$\frac{\frac{9}{17}}{\frac{8}{17}}=\frac{9}{17}\times \frac{8}{17}=\frac{9}{8}$

Probability Exercise 30.5 Question 26

Answer:
$P\left ( A\: winning \right )=\frac{4}{7}$
$P\left ( B\: winning \right )=\frac{2}{7}$
$P\left ( C\: winning \right )=\frac{1}{7}$
Hint: You must know the rules of finding probability functions.
Given: A, B, C are in order. The one to throw a head wins.
Solution: P(A winning) = P(head in first toss)+P(head in 4th toss)+….
$\begin{aligned} &=\frac{1}{2}+\left(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\right)+\ldots \ldots \\ &=\frac{1}{2}\left[1+\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{2}\right)^{6}+\ldots\right] \\ &=\frac{1}{2}\left[\frac{1}{1-\left(\frac{1}{2}\right)^{3}}\right] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \quad\left[1+a+a^{2}+a^{3}+\ldots=\frac{1}{1-a}\right] \\ \end{aligned}$
$\begin{aligned} &=\frac{1}{2}\left[\frac{1}{1-\frac{1}{8}}\right] \\ &=\frac{1}{2} \times \frac{8}{7} \\ &=\frac{4}{7} \end{aligned}$

P(B winning)= P(head in second toss)+P(head in 5th toss)+….

$\begin{aligned} &=\left(\frac{1}{2} \times \frac{1}{2}\right)+\left(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\right)+\ldots \ldots . \\ &=\frac{1}{4}\left[1+\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{2}\right)^{6}+\ldots\right] \\ &=\frac{1}{4}\left[\frac{1}{1-\left(\frac{1}{2}\right)^{3}}\right] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \quad\left[1+a+a^{2}+a^{3}+\ldots=\frac{1}{1-a}\right] \\ \end{aligned}$

$\begin{aligned} &=\frac{1}{4}\left[\frac{1}{1-\frac{1}{8}}\right] \\\\ &=\frac{1}{4} \times \frac{8}{7} \\\\ &=\frac{2}{7} \end{aligned}$

P (C winning)= P(head in third toss)+P(head in 6th toss)+….

$\begin{aligned} &=\left(\frac{1}{2} \times \frac{1}{2}\times \frac{1}{2}\right)+\left(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\times \frac{1}{2}\right)+\ldots \ldots . \\ &=\frac{1}{8}\left[1+\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{2}\right)^{6}+\ldots\right] \\ &=\frac{1}{8}\left[\frac{1}{1-\left(\frac{1}{2}\right)^{3}}\right] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \quad\left[1+a+a^{2}+a^{3}+\ldots=\frac{1}{1-a}\right] \\ \end{aligned}$

$\begin{aligned} &=\frac{1}{8}\left[\frac{1}{1-\frac{1}{8}}\right] \\\\ &=\frac{1}{8} \times \frac{8}{7} \\\\ &=\frac{1}{7} \end{aligned}$

Probability Exercise 30.5 Question 27

Answer:

$P\left ( A\: winning \right )=\frac{36}{91}$
$P\left ( B\: winning \right )=\frac{30}{91}$
$P\left ( C\: winning \right )=\frac{25}{91}$

Hint: You must know the rules of finding probability functions.

Given: Three persons A,B,C throw a die in succession till one gets a six and wins the game.

Solution:

$\begin{aligned} &P(\operatorname{six})=\frac{1}{6} \\ &P(\text { no } \operatorname{six})=\frac{5}{6} \\ &P(\text { A winning })=P(6 \text { in first throw })+P(6 \text { in fourth throw })+\ldots \\ \end{aligned}$

$\begin{aligned} &=\frac{1}{6}+\left(\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}\right)+\ldots \ldots . \\ &=\frac{1}{6}\left[1+\left(\frac{5}{6}\right)^{3}+\left(\frac{5}{6}\right)^{6}+\ldots\right] \\ \end{aligned}$

$\begin{aligned} &=\frac{1}{6}\left[\frac{1}{1-\frac{125}{216}}\right] \quad\left[1+a+a^{2}+a^{3}+\ldots=\frac{1}{1-a}\right] \\\\ &=\frac{1}{6} \times \frac{216}{91} \\\\ &=\frac{36}{91} \end{aligned}$

$\begin{aligned} &P(\mathrm{~B} \text { winning })=P(6 \text { in second throw })+P(6 \text { in fifth throw })+\ldots \\ \end{aligned}$

$\begin{aligned} &=\frac{5}{6} \times \frac{1}{6}+\left(\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}\right)+\ldots . . . \\\\ &=\frac{5}{36}\left[1+\left(\frac{5}{6}\right)^{3}+\left(\frac{5}{6}\right)^{6}+\ldots\right] \\ \end{aligned}$

$\begin{aligned} &=\frac{5}{36}\left[\frac{1}{1-\frac{125}{216}}\right] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \quad\left[1+a+a^{2}+a^{3}+\ldots=\frac{1}{1-a}\right] \\\\ &=\frac{5}{36} \times \frac{216}{91} \\\\ &=\frac{30}{91} \end{aligned}$

$\begin{aligned} &P(\mathrm{~C} \text { winning })=P(6 \text { in third throw })+P(6 \text { in sixth throw })+\ldots \\ \end{aligned}$

$\begin{aligned} &=\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}+\left(\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}\right)+\ldots \ldots \\\\ &=\frac{25}{216}\left[1+\left(\frac{5}{6}\right)^{3}+\left(\frac{5}{6}\right)^{6}+\ldots\right] \\ \end{aligned}$

$\begin{aligned} &=\frac{25}{216}\left[\frac{1}{1-\frac{125}{216}}\right] \quad\left[1+a+a^{2}+a^{3}+\ldots=\frac{1}{1-a}\right] \\\\ &=\frac{25}{216} \times \frac{216}{91} \\\\ &=\frac{25}{91} \end{aligned}$

Probability Exercise 30.5 Question 28

Answer:Proved
Hint: You must know the rules of finding probability functions.
Given: A and B take turns in throwing two dice,
The first throw 10 being awarded the prize.
Solution: There are only three possible cases, where the sum of number obtained after throwing
2 dice is 10, i.e.[(4,6) , (5,5) , (6,4)]
$P( sum \: of \: number\: is 10)=\frac{3}{36}=\frac{1}{12}$
$P( sum \: of \: number \: is \: not\: 10)=1-\frac{1}{12}=\frac{11}{12}$
$P(\mathrm{~A} \: winning )=P(10 \: in \: first \: throw)+P(10 \: in\: third \: throw )+\ldots$
$\begin{aligned} &=\frac{1}{12}+\left(\frac{11}{12} \times \frac{11}{12} \times \frac{1}{12}\right)+\ldots \ldots \\ &=\frac{1}{12}\left[1+\left(\frac{11}{12}\right)^{2}+\left(\frac{11}{12}\right)^{4}+\ldots\right] \\ &=\frac{1}{12}\left[\frac{1}{1-\frac{121}{144}}\right] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \quad\left[1+a+a^{2}+a^{3}+\ldots=\frac{1}{1-a}\right] \\ \end{aligned}$
$\begin{aligned} &=\frac{1}{12} \times \frac{144}{23} \\\\ &=\frac{12}{23} \end{aligned}$
$P(\mathrm{~B} \text { winning })=1-P(\mathrm{~A} \text { winning })=\frac{11}{23}$
Now,
$\frac{P(\mathrm{~A} \text { winning })}{P(\mathrm{~B} \text { winning })}=\frac{\frac{12}{23}}{\frac{11}{23}}=\frac{12}{11}$

Probability Exercise 30.5 Question 29

Answer:$\frac{39}{80}$
Hint: You must know the rules of finding probability functions.
Given:
$Bag-A\rightarrow \left ( 3R+5B \right )$
$Bag-B\rightarrow \left ( 2R+3B \right )$
Solution:
It is given that bag A contains 3 red and 5 black balls and bag B contains (2R,3B)
Total no. of balls in bag A = 8 balls

Total no. of balls in bag B =5 balls

$P\left ( one\: red\: and\: 2\: black \right )=$

$P\left ( 1\: red\: from\: bag\: A\: and\:2\: black\: from\: bagB \right )+P\left ( black\: ball\: from\: bag\: A\: and\: remaining\: from\: B \right )$

$\begin{aligned} &=\left(\frac{3}{8} \times \frac{3}{5} \times \frac{2}{4}\right)+\frac{5}{8} \times \frac{2}{5} \times \frac{3}{4} \times 2 \\ \end{aligned}$

$\begin{aligned} &=\frac{9}{80}+\frac{30}{80}=\frac{39}{80} \end{aligned}$

Probability Exercise 30.5 Question 30 (i)

Answer:$\frac{1}{35}$
Hint: You must know the rules of finding probability functions.
Given:
$P\left ( Fatima\: gets\: selected \right )=\frac{1}{7}=P\left ( A \right )$
$P\left ( John\: gets\: selected \right )=\frac{1}{5}=P\left ( B \right )$
Solution:
$P\left ( Both \: gets\: selected \right )=\frac{1}{5}=P\left ( A\cap B \right )$
$=P\left ( A \right )\times P\left ( B \right )$
$=\frac{1}{7}\times\frac{1}{5} =\frac{1}{35}$

Probability Exercise 30.5 Question 30 (ii)

Answer: $\frac{2}{7}$
Hint: You must know the rules of finding probability functions.
Given:
$P\left ( Fatima\: gets\: selected \right )=\frac{1}{7}=P\left ( A \right )$

$P\left (John\: gets\: selected \right )=\frac{1}{5}=P\left ( B \right )$

Solution:

$P\left (only\:one\: of\: them gets\: selected \right )=P\left ( A \right )\times P\left ( \bar{B} \right )+P\left ( \bar{A} \right )\times P\left ( B \right )$

$\begin{aligned} &=\frac{1}{7}\left(1-\frac{1}{5}\right)+\left(1-\frac{1}{7}\right) \frac{1}{5} \\\\ &=\frac{1}{7} \times \frac{4}{5}+\frac{6}{7} \times \frac{1}{5} \\\\ &=\frac{4}{35}+\frac{6}{35} \\\\ &=\frac{10}{35}=\frac{2}{7} \end{aligned}$

Probability Exercise 30.5 Question 30 (iii)

Answer:$\frac{24}{35}$
Hint: You must know the rules of finding probability functions.
Given:
$P\left ( Fatima\: gets\: selected \right )=\frac{1}{7}=P\left ( A \right )$
$P\left ( John\: gets\: selected \right )=\frac{1}{5}=P\left ( B \right )$
Solution:
$P\left ( none\:one\: of\: them\: gets\: selected \right )=P\left ( \bar{A} \right )\times P\left ( \bar{B} \right )$
$\begin{aligned} &=\left(1-\frac{1}{5}\right) \times\left(1-\frac{1}{7}\right) \\ &=\frac{4}{5} \times \frac{6}{7} \\ &=\frac{24}{35} \end{aligned}$

Probability Exercise 30.5 Question 31 (i)

Answer:$\frac{15}{64}$
Hint: You must know the rules of finding probability functions.
Given: Bag contain 3 blue and 5 red marbles
Solution:
Total no. of marbles =8 marbles
$P\left ( blue\: followed\: by\: red \right )=\frac{3}{8}\times \frac{5}{8}$
$=\frac{15}{64}$

Probability Exercise 30.5 Question 31 (ii)

Answer: $\frac{15}{32}$
Hint: You must know the rules of finding probability functions.
Given: Bag contain 3 blue and 5 red marbles
Solution:
Total no. of marbles =8 marbles
$\begin{aligned} &P(\text { red and blue in any order })=P(\text { blue followed by red })+P(\text { red followed by blue }) \\ \end{aligned}$
$\begin{aligned} &=\frac{3}{8} \times \frac{5}{8}+\frac{5}{8} \times \frac{3}{8} \\\\ &=\frac{15}{64}+\frac{15}{64} \\\\ &=\frac{30}{64}=\frac{15}{32} \end{aligned}$

Probability Exercise 30.5 Question 31 (iii)

Answer:$\frac{17}{32}$
Hint: You must know the rules of finding probability functions.
Given: Bag contain 3 blue and 5 red marbles
Solution:
Total no. of marbles =8 marbles
$\begin{aligned} &P(\text { same colour })=P(\text { both red })+P(\text { both blue }) \\ &=\frac{5}{8} \times \frac{5}{8}+\frac{3}{8} \times \frac{3}{8} \\ &=\frac{25+9}{64} \\ &=\frac{34}{64}=\frac{17}{32} \end{aligned}$

Probability Exercise 30.5 Question 32 (i)

Answer:$\frac{49}{121}$
Hint: You must know the rules of finding probability .
Given: Urn contains 7 red and 4 blue balls.
Two balls are drawn with replacement.
Solution:$P(2 \text { red balls })=\frac{7}{11} \times \frac{7}{11}=\frac{49}{121}$

Probability Exercise 30.5 Question 32 (ii)

Answer: $\frac{16}{121}$
Hint: You must know the rules of finding probability.
Given: Urn contains 7 red and 4 blue balls.
Two balls are drawn with replacement.
Solution:$P(2 \text { blue balls })=\frac{4}{11} \times \frac{4}{11}=\frac{16}{121}$

Probability Exercise 30.5 Question 32 (iii)

Answer: $\frac{56}{121}$
Hint: You must know the rules of finding probability functions.
Given: Urn contains 7 red and 4 blue balls.
Two balls are drawn with replacement
Solution:
$P\left ( 1\: red\: balls \: and\: 1\: blue\: ball\right )=$
$P\left ( blue\: \: balls \: \: followed\: \: by\: \: red\: \: ball\right )+\left ( red\: \: balls \: \: followed\: \: by\: \: blue\: \: ball\right )$
$\begin{aligned} &=\left(\frac{4}{11} \times \frac{7}{11}\right)+\left(\frac{7}{11} \times \frac{4}{11}\right) \\ &=\frac{28}{121}+\frac{28}{121} \\ &=\frac{56}{121} \end{aligned}$

Probability Exercise 30.5 Question 33 (i)

Answer:$\frac{1}{4}$
Hint: You must know the rules of finding probability functions.
Given: Card is drawn from 52 cards, the outcome is noted , then it is replaced and reshuffled,
Another card is drawn.
Solution
We know that there are four suits club (C), spades(S), heart(H), diamond(D), each contain 13 cards.
P(both the cards are of same suit)
$\begin{aligned} &=P\left[\left(C_{1} \cap C_{2}\right) \cup\left(S_{1} \cap S_{2}\right) \cup\left(H_{1} \cap H_{2}\right) \cup\left(D_{1} \cap D_{2}\right)\right] \\\\ &=P\left(C_{1} \cap C_{2}\right)+P\left(S_{1} \cap S_{2}\right)+P\left(H_{1} \cap H_{2}\right)+P\left(D_{1} \cap D_{2}\right) \\\\ &=P\left(C_{1}\right) \times P\left(C_{2}\right)+P\left(S_{1}\right) \times P\left(S_{2}\right)+P\left(H_{1}\right) \times P\left(H_{2}\right)+P\left(D_{1}\right) \times P\left(D_{2}\right) \\\\ &=\frac{13}{52} \times \frac{13}{52}+\frac{13}{52} \times \frac{13}{52}+\frac{13}{52} \times \frac{13}{52}+\frac{13}{52} \times \frac{13}{52} \\ &=4\left(\frac{1}{4} \cdot \frac{1}{4}\right) \\\\ &=\frac{1}{4} \end{aligned}$

Probability Exercise 30.5 Question 33 (ii)

Answer: $\frac{1}{338}$
Hint: You must know the rules of finding probability functions.
Given: Card is drawn from 52 cards, the outcome is noted, the card is replaced and reshuffled,
Another card is drawn.
Solution
We have, four ace and 2 red queens.
P(first card an ace and second card a red queen)
=P(getting an ace)×P(getting a red queen)
$=\frac{4}{52}\times \frac{2}{52}$
$=\frac{1}{13}\times \frac{1}{26}$
$=\frac{1}{338}$

Probability Exercise 30.5 Question 34 (i)

Answer:$\frac{17}{33}$
Hint: You must know the rules of finding probability functions.
Given: Out of 100 students, two section of 40 and 60 are formed
Solution: When both enter the same section.
Here are the possibilities of two cases:
Case-(1)$\rightarrow$enter both are in section A
If both are in section A, 40 students out of 100 can be selected $n(s)={ }^{100} C_{40}$
and (40-2)=38 students out of (100-2)=98
Can be selected $n(E)={ }^{98} C_{38}$
So,
$\begin{aligned} &P(E)=\frac{n(E)}{n(s)}=\frac{{ }^{98} C_{38}}{{ }^{100} C_{40}} \\ &=\frac{98 !}{38 !} \times \frac{40 !}{100 !} \times \frac{60 !}{60 !} \\ &=\frac{1}{100 \times 99} \times 40 \times 39 \\ &=\frac{26}{165} \end{aligned}$
Case-(2)$\rightarrow$if both are in section B, 60 students out of 100 can be selected $n(s)={ }^{100} C_{60}$
and (60-2)=58 students out of (100-2)=98
Can be selected ${ }^{98} C_{58}$
So,
$\begin{aligned} &P(E)=\frac{n(E)}{n(s)}=\frac{{ }^{98} C_{58}}{{ }^{100} C_{60}} \\ &=\frac{98 !}{58 !} \times \frac{60 !}{100 !} \times \frac{40 !}{40 !} \\ &=\left\{\frac{1}{100} \times 99\right\} \times\{60 \times 59\} \times 1 \\ &=\frac{59}{165} \end{aligned}$

Hence, probability that student are either in sector

$\begin{aligned} &=\frac{26}{165}+\frac{59}{165} \\ &=\frac{26+59}{165}=\frac{85}{165}=\frac{17}{33} \end{aligned}$

Probability Exercise 30.5 Question 34 (i)

Hence, probability that student are either in sector

$\begin{aligned} &=\frac{26}{165}+\frac{59}{165} \\ &=\frac{26+59}{165}=\frac{85}{165}=\frac{17}{33} \end{aligned}$

Probability Exercise 30.5 Question 34 (ii)

Answer: $\frac{16}{33}$
Hint: You must know the rules of finding probability functions.
Given: Out of 100 students, two section of 40 and 60 are formed
Solution: We know,
$P\left ( E \right )=1-P\left ( {E}' \right )$
E.g. The probability that both enter different section =
$1-$probability that both enter same section.
$=1-\frac{17}{33}$
$=\frac{33-17}{33}=\frac{16}{33}$

Probability Exercise 30.5 Question 35

Answer:$P\left ( Team\: A \right )=\frac{6}{11}$
$P\left ( Team\: B \right )=\frac{5}{11}$
The decision was fair as the two probabilities are almost equal.
Hint: You must know the rules of finding probability functions.
Given: In a hockey match, both teams scored same number of goals upto end of the game.
Solution:
$P\left ( a\: six \right )=\frac{1}{6}$
$P\left ( not\: a\: six \right )=\frac{5}{6}$
$\begin{aligned} &P(\mathrm{~A} \text { wins })=P(6 \text { in first throw })+P(6 \text { in third throw })+\ldots \\ \end{aligned}$
$\begin{aligned} &=\frac{1}{6}+\left(\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}\right)+\ldots . . \\ &=\frac{1}{6}\left[1+\left(\frac{5}{6}\right)^{2}+\left(\frac{5}{6}\right)^{4}+\ldots\right] \\ \end{aligned}$
$\begin{aligned} &=\frac{1}{6}\left[\frac{1}{1-\frac{25}{36}}\right] \quad\left[1+a+a^{2}+a^{3}+\ldots=\frac{1}{1-a}\right] \\ &=\frac{1}{6} \times \frac{36}{11} \\ &=\frac{6}{11} \end{aligned}$
$\begin{aligned} &P(\mathrm{~B} \text { wins })=P(6 \text { in second throw })+P(6 \text { in fourth throw })+\ldots \\ \end{aligned}$
$\begin{aligned} &=\frac{5}{6} \times \frac{1}{6}+\left(\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}\right)+\ldots . . \\\\ &=\frac{5}{36}\left[1+\left(\frac{5}{6}\right)^{2}+\left(\frac{5}{6}\right)^{4}+\ldots .\right] \\ \end{aligned}$
$\begin{aligned} &=\frac{5}{36}\left[\frac{1}{1-\frac{25}{36}}\right] \quad\left[1+a+a^{2}+a^{3}+\ldots=\frac{1}{1-a}\right] \\\\ &=\frac{5}{36} \times \frac{36}{11} \\\\ &=\frac{5}{11} \end{aligned}$

Probability Exercise 30.5 Question 36

Answer:$\frac{5}{17}$
Hint: $1+a+a^{2}+..........=\frac{1}{1-a}$
Given: A and B throw a pair of dice alternately.
A wins the game if he gets a total of 7 and B wins if he gets 10
Solution: Total of 7 on dice can be obtained in following ways:
$(1,6) , (6,1) , (2,5) , (5,2) , (3,4) , (4,3)$
$Probability\: of\: getting\: 7=\frac{6}{36}=\frac{1}{6}$
$Probability of not getting 7=1-\frac{1}{6}=\frac{5}{6}$
$Total 10 on dice : (4,6),(6,4),(5,5)$
$Probability of getting 10=\frac{3}{36}=\frac{1}{12}$
$Probability of not getting 10=1-\frac{1}{12}=\frac{11}{12}$
$E=getting7,F=getting10$
$P(\mathrm{E})=\frac{1}{6}, P(\overline{\mathrm{E}})=\frac{5}{6}, P(\mathrm{~F})=\frac{1}{12}, P(\overline{\mathrm{F}})=\frac{11}{12}$
Probabilities of getting 7 in first throw$=\frac{1}{6}$
$\begin{aligned} &\therefore P(\text { getting } 7 \text { in third throw })=P(\bar{E}) P(\bar{F}) P(E)=\frac{5}{6} \times \frac{1}{6} \times \frac{11}{12} \\\\ &P(\text { getting } 7 \text { in fifth throw })=P(\bar{E}) P(\bar{F}) P(E)=\frac{5}{6} \times \frac{11}{12} \times \frac{1}{6} \\\\ &P(\text { winning } A)=\frac{1}{6}+\left(\frac{5}{6} \times \frac{11}{12} \times \frac{1}{6}\right)+\left(\frac{5}{6} \times \frac{11}{12} \times \frac{5}{6} \times \frac{11}{12} \times \frac{1}{6}\right)+\ldots . . \end{aligned}$
$\begin{aligned} &\frac{\frac{1}{6}}{1-\frac{5}{6} \times \frac{11}{12}}=\frac{12}{17} \\\\ &\therefore P(\text { winning } B)=1-P(\text { winning } A) \\\\ &=1-\frac{12}{17} \\\\ &=\frac{5}{17} \end{aligned}$

Probability Exercise 30.5 Question 37

Answer:
$P\left ( A\: wins \right )=\frac{6}{11}$
$P\left ( B\: wins \right )=\frac{5}{11}$

Hint: You must know the rules of finding probability functions.

cgfyTill one of them gets the sum of numbers as multiples of 6 and wins the game.

Solution: Let S denote the success and F denote the failure (not getting 6)

Thus,

$\begin{aligned} &P(\mathrm{~S})=\frac{1}{6}=\mathrm{p} \quad, \quad P(\mathrm{~F})=\frac{5}{6}=q \\\\ &P(\mathrm{~A} \text { wins the first throw })=P(\mathrm{~S})=p \\\\ &P(\mathrm{~A} \text { wins the third throw })=P(\mathrm{FFS})=q q p \\\\ &P(\mathrm{~A} \text { wins the fifth throw })=P(\mathrm{FFFFS})=q q q q p \end{aligned}$

So,

$\begin{aligned} &P(\mathrm{~A} \text { wins })=p+q q p+q q q q p+\ldots \\\\ &=p\left(1+q^{2}+q^{4}+\ldots\right) \\\\ &=\frac{p}{1-q^{2}}=\frac{\frac{1}{6}}{1-\frac{25}{36}}=\frac{6}{11} \\\\ &P(\mathrm{~B} \text { wins })=1-\frac{6}{11}=\frac{5}{11} \\\\ &\therefore P(\mathrm{~A} \text { wins })=\frac{6}{11} \quad, \quad P(\mathrm{~B} \text { wins })=\frac{5}{11} \end{aligned}$

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