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  • Provide solution for RD Sharma Maths Class 12 Chapter 30 Probability Exercise 30.3 question 10.

Provide solution for RD Sharma Maths Class 12 Chapter 30 Probability Exercise 30.3 question 10.

Answers (1)

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}           

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