RD Sharma Class 12 Exercise 30.3 Probability Solutions Maths-Download PDF Online

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RD Sharma Class 12 Solutions Chapter30 Probability - Other Exercise
Probability Excercise: 30.3
RD Sharma Chapter-wise Solutions

Probability Excercise: 30.3

Probability Exercise 30.3 Question 1

Answer: $\frac{4}{9}$
Hint: Just simplify and use formula of probability
Given:
$\begin{aligned} &P(A)=\frac{7}{13} \\ &P(B)=\frac{9}{13} \\ &P(A \cap B)=\frac{4}{13} \end{aligned}$
Solution:
Using, $P(A \mid B)=\frac{P(A \cap B)}{P(B)} \text { and } P(B \mid A)=\frac{P(A \cap B)}{P(A)}$
Required Probability,
$P(A \mid B)=\frac{\frac{4}{18}}{\frac{9}{18}}=\frac{4}{9}$

Probability Exercise 30.3 Question 2

Answer: $\frac{2}{3}, \frac{1}{3}$
Hint: $\text{Probability}=\frac{\text{No. of outcomes}}{\text{Total number of outcomes}}$
Given:
$\begin{aligned} &P(A)=0.6 \\ &P(B)=0.3 \\ &P(A \cap B)=0.2 \end{aligned}$
Solution:
Now,
$\begin{aligned} &P(A \mid B)=\frac{P(A \cap B)}{P(B)} \\ &=\frac{0.2}{0.3} \\ &=\frac{2}{3} \\ &P(B \mid A)=\frac{P(A \cap B)}{P(A)} \\ &=\frac{0.2}{0.6} \\ &=\frac{1}{3} \end{aligned}$

Probability Exercise 30.3 Question 3

Answer: 0.64
Hint: $P(A \mid B)=\frac{P(A \cap B)}{P(B)}$
Given:
$\begin{aligned} &P(B)=0.5 \\ &P(A \cap B)=0.32 \end{aligned}$
Solution:
$\begin{aligned} &P(A \mid B)=\frac{P(A \cap B)}{P(B)} \\ &=\frac{0.32}{0.5} \\ &=0.64 \end{aligned}$

Probability Exercise 30.3 Question 4

Answer: 0.96
Hint: Use $P(A \cup B)=P(A)+P(B)-P(A \cap B)$
Given,
$\begin{aligned} &P(A)=0.4 \\ &P(B)=0.8 \\ &P\left(\frac{B}{A}\right)=0.6 \end{aligned}$
Solution:
Now,
$\begin{aligned} &P(A \mid B)=\frac{P(A \cap B)}{P(B)} \\ &P(A \cap B)=0.24 \\ &\Rightarrow P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)} \\ &=\frac{0.24}{0.8} \\ &=0.3 \\ &\Rightarrow P(A \cup B)=P(A)+P(B)-P(A \cap B) 4 \\ &=0.4+0.8-0.24 \\ &=0.96 \end{aligned}$

Probability Exercise 30.3 Question 5(i)

Answer: $\frac{2}{3},\frac{1}{2}$
Hint:
$\begin{aligned} &P(A \mid B)=\frac{P(A \cap B)}{P(B)} \\ &P(B \mid A)=\frac{P(A \cap B)}{P(A)} \end{aligned}$
Given,
$\begin{aligned} &P(A)=\frac{1}{3} \\ &P(B)=\frac{1}{4} \\ &P(A \cup B)=\frac{5}{12} \end{aligned}$
$\begin{aligned} &P(A \cup B)=P(A)+P(B)-P(A \cap B) \\ &P(A \cap B)=P(A)+P(B)-P(A \cup B) \end{aligned}$
$\begin{aligned} &P(A \cap B)=\frac{1}{3}+\frac{1}{4}-\frac{5}{12} \\ &P(A \cap B)=\frac{2}{12} \\ &P(A \cap B)=\frac{1}{6} \end{aligned}$
Solution:
Now,
$\begin{gathered} P(A \mid B)=\frac{P(A \cap B)}{p(B)} \\ =\frac{\frac{1}{6}}{\frac{1}{4}} \\ =\frac{4}{6}=\frac{2}{3} \end{gathered}$
$\begin{gathered} P(B \mid A)=\frac{P(A \cap B)}{P(A)} \\ =\frac{\frac{1}{6}}{\frac{1}{3}} \\ =\frac{1}{2} \end{gathered}$

Probability Exercise 30.3 Question 5(ii)

Answer: $\frac{4}{5},\frac{2}{3}$
Hint:
$\begin{aligned} &\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\ &\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cup \mathrm{B}) \end{aligned}$
Solution:
Given,
$\begin{aligned} &P(A)=\frac{6}{11} \\ &P(B)=\frac{5}{11} \\ &P(A \cup B)=\frac{7}{11} \end{aligned}$
Now,
$\begin{aligned} &P(A \mid B)=\frac{P(A \cap B)}{P(B)} \\ &=\frac{\frac{4}{11}}{\frac{5}{11}} \\ &=\frac{4}{5} \end{aligned}$
$\begin{aligned} P(B \mid A) &=\frac{P(A \cap B)}{P(A)} \\ &=\frac{\frac{4}{11}}{\frac{6}{11}} \end{aligned}$
$=\frac{4}{6}$ $=\frac{2}{3}$

Probability Exercise 30.3 Question 5(iii)

Answer: $\frac{5}{9}$

Hint: $P(\bar{A} \cap B)=P(B)-P(A \cap B)$
Given:
$\begin{aligned} &P(A)=\frac{7}{13} \\ &P(B)=\frac{9}{13} \\ &P(A \cap B)=\frac{4}{13} \end{aligned}$
Solution:
$\begin{aligned} &P(\bar{A} \cap B)=P(B)-P(A \cap B) \\ &=\frac{9}{13}-\frac{4}{13} \\ &=\frac{5}{13} \end{aligned}$
Now,
$\begin{aligned} &P(\bar{A} / B)=\frac{P(\bar{A} \cap B)}{P(B)} \\ &=\frac{\frac{5}{13}}{\frac{9}{13}} \\ &=\frac{5}{9} \end{aligned}$

Probability Exercise 30.3 Question 5(iv)

Answer: $\frac{1}{4}, \frac{5}{8}$
Hint: $P(A \cup B)=P(A)+P(B)-P(A \cap B)$
Given
$\begin{aligned} &P(A)=\frac{1}{2} \\ &P(B)=\frac{1}{3} \end{aligned}$
Solution:
$P(A \cap B)=\frac{1}{4}$
Also,
$\begin{aligned} &\mathrm{P}(\overline{\mathrm{B}})=1-\mathrm{P}(\mathrm{B}) \\ &=1-\frac{1}{3} \\ &=\frac{2}{3} \\ &\Rightarrow \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\ &=\frac{1}{2}+\frac{1}{3}-\frac{1}{4} \\ &=\frac{6+4-3}{12} \\ &=\frac{7}{12} \end{aligned}$
Also,
$\begin{aligned} &P(\bar{A} \cap B)=P(B)-P(A \cap B) \\ &=\frac{1}{3}-\frac{1}{4} \\ &=\frac{4-3}{12} \\ &P(\bar{A} \cap B)=\frac{1}{12} \end{aligned}$
Solution:
We know
$\begin{aligned} &P(\bar{A} \cap \bar{B})=P(\overline{A \cup B}) \\ &=1-P(A \cup B) \\ &=1-\frac{7}{12} \\ &=\frac{5}{12} \end{aligned}$
$\begin{aligned} &P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)} \\ &=\frac{\frac{1}{4}}{\frac{1}{3}} \\ &=\frac{3}{4} \end{aligned}$
$\begin{aligned} &P(B \mid A)=\frac{P(A \cap B)}{P(A)} \\ &=\frac{\frac{1}{4}}{\frac{1}{2}} \\ &=\frac{2}{4}=\frac{1}{2} \end{aligned}$
$\begin{aligned} &P\left(\frac{\bar{A}}{B}\right)=\frac{P(\bar{A} \cap B)}{P(B)} \\ &=\frac{\frac{1}{12}}{\frac{1}{3}} \\ &=\frac{3}{12}=\frac{1}{4} \\ &P\left(\frac{\bar{A}}{\bar{B}}\right)=\frac{P(\bar{A} \cap \bar{B})}{P(\bar{B})} \end{aligned}$
$\begin{aligned} &=\frac{\frac{5}{12}}{\frac{2}{3}} \\ &=\frac{15}{24}=\frac{5}{8} \end{aligned}$

Probability Exercise 30.3 Question 6

Answer: $\frac{11}{26}$
Hint: $P(A \cup B)=P(A)+P(B)-P(A \cap B)$
Given,
$\begin{aligned} &2 P(A)=P(B)=\frac{5}{13} \\ &P(A \mid B)=\frac{2}{5} \\ &P(A)=\frac{5}{26} \\ &P(B)=\frac{5}{13} \end{aligned}$
Solution:
Now,
$\begin{aligned} &P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)} \\ &\frac{2}{5}=\frac{P(A \cap B)}{\frac{5}{13}} \\ &P(A \cap B)=\frac{2}{5} \times \frac{5}{13} \end{aligned}$
$=\frac{2}{13}$
$\Rightarrow P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$\begin{aligned} &=\frac{5}{26}+\frac{5}{13}-\frac{2}{13} \\ &=\frac{11}{26} \end{aligned}$

Probability Exercise 30.3 Question 7

Answer: $\frac{4}{11},\frac{4}{5},\frac{2}{3}$
Hint: Use $\Rightarrow P(A \cup B)=P(A)+P(B)-P(A \cap B)$
Given,
$\begin{aligned} &P(A)=\frac{6}{11} \\ &P(B)=\frac{5}{11} \\ &P(A \cup B)=\frac{7}{11} \end{aligned}$
Solution:
Now, (i)
$\begin{aligned} &P(A \cup B)=P(A)+P(B)-P(A \cap B) \\ &\frac{7}{11}=\frac{6}{11}+\frac{5}{11}-P(A \cap B) \\ &P(A \cap B)=1-\frac{7}{11} \\ &P(A \cap B)=\frac{4}{11} \end{aligned}$

(ii)
$\begin{aligned} &P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)} \\ &=\frac{\frac{4}{11}}{\frac{5}{11}} \\ &=\frac{4}{5} \end{aligned}$
(iii)
$\begin{aligned} &P(B \mid A)=\frac{P(A \cap B)}{P(A)} \\ &=\frac{\frac{4}{11}}{\frac{6}{11}} \\ &=\frac{4}{6}=\frac{2}{3} \end{aligned}$

Probability Exercise 30.3 Question 8(i)

Answer: $\frac{1}{2}$
Hint: $\text{Probability}=\frac{\text{No. of Outcomes}}{\text{Total no of outcomes}}$
Given, A=Heads on third toss
B=Heads on first two tosses
Solution:
$\begin{aligned} &S=\{(H H H),(H H T),(H H T),(T T H),(T T T),(H T H),(T H H),(T T H)\} \\ &A=\{(H H H),(H T H),(T H H),(T T H)\} \\ &B=\{(H H H),(H H T)\} \\ &A \cap B=\{H H H\} \end{aligned}$
$P(A)=4, P(B)=2, P(A \cap B)=1$
Required Probability $=P\left (\frac{A}{B} \right )$
$\begin{aligned} &=\frac{P(A \cap B)}{P(B)} \\ &=\frac{1}{2} \end{aligned}$

Probability Exercise 30.3 Question 8(ii)

Answer: $=P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}=\frac{3}{7}$
Hint: $\text{Probability}=\frac{\text{No. of outcomes}}{\text{Total no.of Outcomes }}$
Given, A=At least two heads
B=At most two heads
Solution:
Clearly,
$S=\{(H H H),(H H T),(H H T),(T T H),(T T T),(H T H),(T H H),(T T H)\}$
$\begin{aligned} &\mathrm{A}=\{(\mathrm{HHH}),(\mathrm{HTH}),(\mathrm{THH}),(\mathrm{HHT})\} \\ &\mathrm{B}=\{(\mathrm{TT}),(\mathrm{HTH}),(\mathrm{HTT}),(\mathrm{THT}),(\mathrm{THH}),(\mathrm{HHT}),(\mathrm{HTT})\} \\ &\mathrm{A} \cap \mathrm{B}=\{(\mathrm{HTH}),(\mathrm{THH}),(\mathrm{HHT})\} \end{aligned}$
$P(A)=4$
$P(B)=7$
Required Probability $=P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}=\frac{3}{7}$

Probability Exercise 30.3 Question 8(iii)

Answer : $=P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}=\frac{6}{7}$
Hint: $\text{Probability}=\frac{\text{No. of outcomes}}{\text{Total no.of Outcomes }}$
Given,
A=At most two tails
B=At least one tail
Solution:
Clearly,
$\begin{aligned} &S=\{(H H H),(H H T),(H H T),(T T H),(T T T),(H T H),(T H H),(T T H)\} \\ &A=\{(\pi H),(T H H),(\mathrm{HHT}),(T H T),(\mathrm{HHT}),(\mathrm{HTT}),(\mathrm{HHH})\} \\ &\mathrm{B}=\{(\mathrm{TT}),(\mathrm{TH}),(\mathrm{THH}),(\mathrm{HHT}),(\mathrm{THT}),(\mathrm{HHT}),(\mathrm{HTT})\} \end{aligned}$
Now, $\mathrm{A} \cap \mathrm{B}=\{(\mathrm{TTH}),(\mathrm{THH}),(\mathrm{HHT}),(\mathrm{HTT})\}$
P(A) = 7
P(B) = 7
Required Probability $=P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}=\frac{6}{7}$

Probability Exercise 30.3 Question 9(i)

Answer: 1
Hint: $\text{Probability}=\frac{\text{No. of outcomes}}{\text{Total no.of Outcomes }}$
Given, A=Tail appears on one coin
B=One coin shows head
Solution:
Clearly,
$\begin{aligned} &\mathrm{A}=\{(\mathrm{H}, \mathrm{T}),(\mathrm{T}, \mathrm{H})\} \\ &\mathrm{B}=\{(\mathrm{H}, \mathrm{T}),(\mathrm{T}, \mathrm{H})\} \end{aligned}$
P(A) = 2
P(B) = 2
Now, $\mathrm{A} \cap \mathrm{B}=\{(\mathrm{H}, \mathrm{T}),(\mathrm{T}, \mathrm{H})\}$

Required Probability $=P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}=\frac{2}{2}=1$

Probability Exercise 30.3 Question 9(ii)

Answer: 0
Hint: $\text{Probability}=\frac{\text{No. of outcomes}}{\text{Total no.of Outcomes }}$
Given events,
A=No tail appears
B=No head appears
Solution:
Clearly,
$\begin{aligned} &A=\{(H, H)\} \\ &B=\{(T, T)\} \end{aligned}$
$\begin{aligned} &A \cap B=\{\phi\} \\ &P(A)=2 \\ &P(B)=2 \end{aligned}$
Required Probability $=P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}=0$

Probability Exercise 30.3 Question 10

Answer: $\frac{1}{6},\frac{1}{36}$
Hint: $\text{Probability}=\frac{\text{No. of outcomes}}{\text{Total no.of Outcomes }}$
Given,
A=Getting 4 on trial throw
B=Getting 6 on first throw and 5 on second throw
Solution:
Clearly,
$\begin{gathered} \mathrm{A}=\left\{\begin{array}{l} (1,1,4),(1,2,4),(1,3,4),(1,4,4),(1,5,4), \\ (1,6,4),(2,1,4),(2,2,4),(2,3,4),(2,4,4), \\ (2,5,4),(2,6,4),(3,1,4),(3,2,4),(3,3,4), \\ (3,4,4),(3,5,4),(3,6,4),(4,1,4),(4,2,4), \\ (4,3,4),(4,4,4),(4,5,4),(4,6,4),(5,1,4), \\ (5,2,4),(5,3,4),(5,4,4),(5,5,4),(5,6,4), \\ (6,1,4),(6,2,4),(6,3,4),(6,4,4),(6,5,4), \\ (6,6,4) \end{array}\right\} \\ \mathrm{B}=\{(6,5,1),(6,5,2),(6,5,3),(6,5,4),(6,5,5),(6,5,6)\} \end{gathered}$

Now,
$\mathrm{A} \cap \mathrm{B}=\{(6,5,4)\}$
P(A) = 36
P(B) = 6
$\begin{aligned} &P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{1}{6} \\ &P(B \mid A)=\frac{P(A \cap B)}{P(A)}=\frac{1}{36} \end{aligned}$

Probability Exercise 30.3 Question 11

Answer: $1,\frac{1}{2}$
Hint: $\text{Probability}=\frac{\text{No. of outcomes}}{\text{Total no.of Outcomes }}$
Given, A= Son standing on one end
B= Father standing in the middle
Solution:
Clearly,
S=Total events
$\begin{aligned} &=\{\mathrm{MFS}, \mathrm{MSF}, \mathrm{FSM}, \mathrm{FMS}, \mathrm{SMF}, \mathrm{SFM}\} \\ &\mathrm{A}=\{\mathrm{MFS}, \mathrm{FMS}, \mathrm{SMF}, \mathrm{SFM}\} \\ &\mathrm{B}=\{\mathrm{MFS}, \mathrm{SFM}\} \\ &\mathrm{A} \cap \mathrm{B}=\{\mathrm{MFS}, \mathrm{SFM}\} \end{aligned}$
$(1)$
$\begin{gathered} P\left(\frac{A}{B}\right)=\frac{n(A \cap B)}{n(B)} \\ =\frac{2}{2} \\ =1 \end{gathered}$
$(2)$
$\begin{gathered} P(B \mid A)=\frac{P(A \cap B)}{P(A)} \\ =\frac{2}{4} \\ =\frac{1}{2} \end{gathered}$

Probability Exercise 30.3 Question 12

Answer: $\frac{2}{5}$
Hint: $\text{Probability}=\frac{\text{No. of outcomes}}{\text{Total no.of Outcomes }}$
Given,
A= 4 appears on the die at least once
B=The sum of the numbers on two dice is 6
Solution:
$\begin{aligned} &\mathrm{A}=\left\{\begin{array}{l} (1,4),(2,4),(3,4),(4,1),(4,2),(4,3), \\ (4,4),(4,5),(4,6),(5,4),(6,4) \end{array}\right\} \\ &\mathrm{B}=\{(1,5),(5,1),(2,4),(4,2),(3,3)\} \\ &\mathrm{A} \cap \mathrm{B}=\{(2,4),(4,2)\} \end{aligned}$
P(A) = 11
P(B) = 5
Required Probability
$\begin{aligned} &=P\left(\frac{A}{B}\right) \\ &=\frac{n(A \cap B)}{n(B)}=\frac{2}{5} \end{aligned}$

Probability Exercise 30.3 Question 13

Answer: $\frac{1}{6}$
Hint: $\text{Probability}=\frac{\text{No. of outcomes}}{\text{Total no.of Outcomes }}$

Solution:
Clearly,
$\begin{aligned} &A=\left\{\begin{array}{l} (1,4),(2,4),(3,4), \\ (4,4),(5,4),(6,4) \end{array}\right\} \\ &B=\{(4,4),(3,5),(5,3),(2,6),(6,2)\} \\ &A \cap B=\{(4,4)\} \end{aligned}$

Required Probability = $P(B \mid A)=\frac{P(A \cap B)}{P(A)}=\frac{1}{6}$

Probability Exercise 30.3 Question 14

Answer: $=P(B \mid A)=\frac{P(A \cap B)}{p(A)}=\frac{1}{6}$
Hint: $\text{Probability}=\frac{\text{No. of outcomes}}{\text{Total no.of Outcomes }}$
Given,
A=No. appearing on second die is odd.
B=The sum of the no. on two dice is 7.
Solution:
Clearly,
$\begin{aligned} &A=\left\{\begin{array}{l} (1,1),(1,3),(1,5),(2,1),(2,3),(2,5), \\ (3,1),(3,3),(3,5),(4,1),(4,3),(4,5) \\ (5,1),(5,3),(5,5),(6,1),(6,3),(6,5) \end{array}\right\} \\ &B=\{(2,5),(5,2),(3,4),(4,3),(1,6),(6,1)\} \\ &A \cap B=\{(2,5),(4,3),(6,1)\} \end{aligned}$
Required Probability $=P(B \mid A)=\frac{P(A \cap B)}{p(A)}=\frac{1}{6}$

Probability Exercise 30.3 Question 15

Answer: $P(B \mid A)=\frac{P(A \cap B)}{p(A)}=\frac{1}{6}$

Hint: Use, $P(B \mid A)=\frac{P(A \cap B)}{p(A)}$
Given,
A= Prime number appears on second die
B=The sum of the numbers on two dice is 7
Solution:
Clearly,
$\begin{aligned} &A=\left\{\begin{array}{l} (1,2),(1,3),(1,5),(2,2),(2,3),(2,5), \\ (3,2),(3,3),(3,5),(4,2),(4,3),(4,5), \\ (5,2),(5,3),(5,5),(6,2),(6,3),(6,5) \end{array}\right\} \\ &B=\{(2,5),(5,2),(3,4),(4,3),(1,6),(6,1)\} \\ &A \cap B=\{(2,5),(5,2),(4,3)\} \end{aligned}$
P(A) = 18
P(B) = 6
Required Probability $P(B \mid A)=\frac{P(A \cap B)}{p(A)}=\frac{1}{6}$

Probability Exercise 30.3 Question 16

Answer: $\frac{2}{3}$
Hint: Use, $P(B \mid A)=\frac{P(A \cap B)}{P(A)}$
Given,
A= No. is odd
B=No. is prime
Solution:
Clearly,
$\begin{aligned} &A=\{1,3,5\} \\ &B=\{2,3,5\} \\ &A \cap B=\{3,5\} \end{aligned}$

Required Probability $P(B \mid A)=\frac{P(A \cap B)}{P(A)}$
$=\frac{2}{3}$

Probability Exercise 30.3 Question 17

Answer: $P(B \mid A)=\frac{n(A \mid B)}{n(A)}=\frac{3}{6}=\frac{1}{2}$
Hint: Use, $P(B \mid A)=\frac{P(A \mid B)}{P(A)}$

Given,
A = 4 appears on first die
B = The sum of the numbers is on two dice is 8 or more
Solution:
$\begin{aligned} &A=\{(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\} \\ &n(A)=6 \\ &B=\left\{\begin{array}{l} (2,6),(3,5),(3,6),(4,4),(4,5), \\ (4,6),(5,3),(5,4),(5,6),(6,2), \\ (6,3),(6,4),(6,5),(6,6) \end{array}\right\} \\ &n(B)=15 \end{aligned}$
Now,
$\begin{aligned} &P(A)=6 \\ &P(B)=15 \\ &A \cap B=\{(4,4),(4,5),(4,6)\} \end{aligned}$
Required Probability $P(B \mid A)=\frac{P(A \mid B)}{P(A)}=\frac{1}{2}$

Probability Exercise 30.3 Question 18

Answer: $\frac{3}{25}$
Hint: $P(B \mid A)=\frac{P(A \cap B)}{P(A)}$
Given,
A=At least one die does not show 5
B=The sum of the numbers on two dice is 8
Solution:
Clearly,

$\begin{aligned} &A=\left\{\begin{array}{l} (1,1),(1,2),(1,3),(1,4),(1,6), \\ (2,1),(2,2),(2,3),(2,4),(2,6), \\ (3,1),(3,2),(3,3),(3,4),(3,6), \\ (4,1),(4,2),(4,3),(4,4),(4,6), \\ (6,1),(6,2),(6,3),(6,4),(6,6) \end{array}\right\} \\ &B=\{(2,6),(3,5),(4,4),(5,3),(6,2)\} \\ &A \cap B=\{(4,4),(6,2),(2,6)\} \end{aligned}$
Required Probability $=P(B \mid A)=\frac{P(A \cap B)}{P(A)}=\frac{3}{25}$

Probability Exercise 30.3 Question 19

Answer: $\frac{10}{5}=\frac{5}{8}$
Hint: Use combination method of $n_{C_{r}}=\frac{n !}{(n-r) ! r !}$
Given: O represents the event of getting two odd numbers
S represents the event of getting their sum as an even number.
Solution:
$\begin{gathered} \mathrm{P}(\mathrm{S})=\left(4 \mathrm{C}_{2}+5 \mathrm{C}_{2}\right) / 9 \mathrm{C}_{2} \\ P(O \mid S)=\frac{n(O \cap S)}{n(S)} \\ =\frac{\frac{\mathrm{s}_{2}}{9} \mathrm{C}_{2}}{\frac{\left({ }^{4} \mathrm{C}_{2}+5_{2}\right)}{{ }^{9} \mathrm{C}_{2}}} \end{gathered}$
$=\frac{5 c_{2}}{\left(4 c_{2}+5_{c 2}\right)} \: \: \: \: \: \: \: \: \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left\{\left(5_{C_{2}}\right)=\frac{5 \times 4 \times 3 \times 2 \times 1}{(5-2) !(2 \times 1)}=\frac{120}{3 \times 2 \times 1 \times 2}=10\right\}$
$\frac{10}{5}=\frac{5}{8}$

Probability Exercise 30.3 Question 20

Answer: $\frac{2}{5}$
Hint: $P(A \mid B)=\frac{n(A \cap B)}{n(B)}$
Given :- Now, Let A = 5 appears on the die at least once.
B = The sum of the no. on two dice is 8
Solution:
Clearly,
$\begin{aligned} &A=\left\{\begin{array}{l} (1,5),(2,5),(3,5),(4,5),(5,5), \\ (6,5),(5,1),(5,2),(5,3),(5,4), \\ (5,6) \end{array}\right\} \\ &B=\{(2,6),(3,5),(4,4),(5,3),(6,2)\} \\ &A \cap B=\{(3,5),(5,3)\} \end{aligned}$

P(A) = 11
P(B) = 5
Required Probability $P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{2}{3}$

Probability Exercise 30.3 Question 21

Answer: $=\frac{1}{6}$
Hint: $=P(B \mid A)=\frac{P(A \cap B)}{p(A)}$

Given :- Let, A=First die shows
B=The sum of the numbers on two dice is 7
Solution:
Clearly,
$\begin{aligned} &A=\{(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\} \\ &B=\{(2,5),(5,2),(4,3),(3,4),(1,6),(6,1)\} \\ &A \cap B=\{(6,1)\} \end{aligned}$
P(A) = 6
P(B) = 6
Required Probability $=P(B \mid A)=\frac{P(A \cap B)}{p(A)}=\frac{1}{6}$

Probability Exercise 30.3 Question 22

Answer: $P(E \mid F)=\frac{P(E) F J}{P(F)}=\frac{2}{6}=\frac{1}{3}$
Hint: $P(B \mid A)=\frac{P(A \cap B)}{P(A)}$
Given, E = The sum of the no. on two dice is 10 or more.
F = 5 appears on first die
Solution:
Clearly,
$\begin{aligned} &E=\{(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)\} \\ &F=\{(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\} \\ &E \cap F=\{(5,5),(5,6)\} \end{aligned}$
P(F) = 6
P(E) = 6
$=P(E \mid F)=\frac{p(E \cap F)}{p(F)}=\frac{2}{6}=\frac{1}{3}$
Second case:
Let, E = The sum of the no. on two dice is 10 or more.
F = 5 appears on at least in one die
Now, P(F) = 11
P(E) = 6
$\begin{aligned} &\mathrm{E}=\{(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)\} \\ &\mathrm{F}=\left\{\begin{array}{l} (1,5),(2,5),(3,5),(4,5),(6,5) \\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \end{array}\right\} \\ &\mathrm{E} \cap \mathrm{F}=\{(5,5),(5,6),(6,5)\} \\ &P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{3}{11} \end{aligned}$

Probability Exercise 30.3 Question 23

Answer: $\frac{5}{8}$
Hint: $P(E \mid F)=\frac{P(E \cap F)}{P(F)}$
Given :- Let, M= Students passes Mathematics,
C=Students Passes Computer Science
$\begin{aligned} &P(M)=\frac{4}{5} \\ &P(M \cap C)=\frac{1}{2} \end{aligned}$
Now,
$\begin{gathered} P(C \mid M)=\frac{P(\mathrm{CnM}]}{p(M)} \\ =\frac{\frac{1}{2}}{\frac{4}{5}} \\ =\frac{5}{8} \end{gathered}$

Probability Exercise 30.3 Question 24

Answer: 0.6
Hint: $P(E \mid F)=\frac{P(E \cap F)}{p(F)}$
Given, S=Person buying a shirt
T=Person buying a trouser
Solution:
$\begin{aligned} &P(S)=0.2 \\ &P(T)=0.3 \\ &P(S \mid T)=0.4 \\ &P(S \mid T)=\frac{P(S \cap T)}{P(T)} \\ &P(S \cap T)=P(S \mid T) \cdot P(T) \\ &=0.4 \times 0.3 \\ &=0.12 \end{aligned}$
Now,
$\begin{aligned} &P(T \mid S)=\frac{P(S \cap T)}{P(S)} \\ &=\frac{0.12}{0.2}=0.6 \end{aligned}$

Probability Exercise 30.3 Question 25

Answer: $\frac{1}{10}$
Hint: $P\left(\frac{S}{G}\right)=\frac{P(S \cap G)}{P(G)}$
Given, S= Student chosen randomly studies in class XII
G= Female student chosen randomly
$\begin{aligned} &P(G)=\frac{430}{1000} \\ &P(S \cap G)=\frac{43}{1000} \\ &P\left(\frac{S}{G}\right)=\frac{P(S \cap G)}{P(G)} \end{aligned}$
$\begin{aligned} =\frac{ \frac{430}{1000}}{ \frac{430}{1000}} \\ = \frac{1}{10} \end{aligned}$

Probability Exercise 30.3 Question 26

Answer: $\frac{4}{7}$
Hint: $\text{Probability}=\frac{\text{No. of outcomes}}{\text{Total no.of Outcomes }}$
Given:
Total Possibility of cards = {1,2,3,4,5,6,7,8,9,10}
Solution:
Let A = An even number on the card
B = A number more than 3 on the card.
$\begin{aligned} &A=\{2,4,6,8,10\} \\ &B=\{4,5,6,7,8,9,10\} \\ &A \cap B=\{4,6,8,10\} \end{aligned}$
P(A) = {5}
P(B) = {7}

Required probability $P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}=\frac{4}{7}$

Probability Exercise 30.3 Question 27

Answer: $\frac{1}{2},\frac{1}{3}$
Hint: $P(A \mid B)=\frac{P(A \cap B)}{p(B)}$
Given, A=Both the children are girls
B=The youngest child is a girl
C=At least one child is a girl
Solution:
$\begin{aligned} &S=\left\{B_{1} B_{2}, B_{1} G_{2}, G_{1} B_{2}, G_{1} G_{2}\right\} \\ &A=\left\{G_{1} G_{2}, G_{2} G_{1}\right\} \\ &B=\left\{B_{1} G_{2}, G_{1} G_{2}\right\} \\ &C=\left\{B_{1} G_{2}, G_{1} B_{2}, G_{1} G_{2}\right\} \\ &A \cap B=\left\{G_{1} G_{2}\right\} \\ &A \cap C=\left\{G_{1} G_{2}\right\} \\ &P(A)=2 \\ &P(B)=2 \end{aligned}$

Required probability $P(A \mid B)=\frac{P(A \cap B)}{p(B)}=\frac{1}{2}$
Required probability $P(A \mid C)=\frac{P(A \cap C)}{p(C)}=\frac{1}{3}$

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