Question

$A$ die is thrown three times. Events $A$ and $B$ are defined as below :
$A{:}^{\prime }4$ on the third throw' and $B{:}^{\prime }6$ on the first and $5$ on the second throw'. Find the probability of $A$ given that $B$ has already occurred.

• A. $\frac{2}{3}$
• B. $\frac{1}{6}$
• C. $\frac{1}{7}$
• D. $\frac{2}{7}$

Answer 1

Answer: (B)

In this case, the sample space is
$S=\{(x,y,z):x,y,z\in \{1,2,3,4,5,6\}\}$, which contains $6\times 6\times 6=216$ equally likely sample points.
Here, $A=\{(x,y,z):z=4$,
$x\in \{1,2,3,4,5,6\}$,
$y\in \{1,2,3,4,5,6\}\}$
and $B=\{(x,y,z):x=6,y=5,z\in \{1,2,3,4,5,6\}\}$
$\Rightarrow A\cap B=\left\{\left(x,y,z\right):x=6,y=5,z=4\right\}=\left\{\left(6,5,4\right)\right\}$
Obviously, ${}^{\prime }{A}^{\prime }$ contains $6\times 6=36$ sample points, ${}^{\prime }{B}^{\prime }$ contains $6$ sample points and $A\cap B$ contains only one sample point.
$\therefore P\left(A/B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}$
$=\frac{\frac{1}{216}}{\frac{6}{216}}=\frac{1}{6}$