Question

A die is thrown three times. Events $A$ and $B$ are defined as below:
$A:4$ on the third throw
$B:6$ on the first and $5$ on the second throw.
The probability of occurrence of $A$ given that $B$ has already occurred is:

• A. $\frac{1}{6}$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. $\frac{1}{5}$

Answer 1

The correct option is A $\frac{1}{6}$

The sample space has $216$ outcomes.
Now $A=\left\{\begin{array}{c}(1,1,4),(1,2,4),...,(1,6,4),(2,1,4),(2,2,4),...,(2,6,4),\\ (3,1,4),(3,2,4),...,(3,6,4),(4,1,4),(4,2,4),...,(4,6,4),\\ (5,1,4),(5,2,4),...,(5,6,4),(6,1,4),(6,2,4),...,(6,6,4)\end{array}\right\}$
$B=\left\{\begin{array}{c}(6,5,1),(6,5,2),(6,5,3),(6,5,4),(6,5,5),(6,5,6)\end{array}\right\}$
and $A\cap B=\left\{\right(6,5,4\left)\right\}$
Now $P\left(B\right)=\frac{6}{216}$ and $P(A\cap B)=\frac{1}{216}$
Then $P\left(A/B\right)=\frac{P(A\cap B)}{P\left(B\right)}=\frac{\frac{1}{216}}{\frac{6}{216}}=\frac{1}{6}$