Question

$A$ die is thrown three times. Events $A$ and $B$ are defined as below :
$A : '4$ on the third throw' and $B : '6$ 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, $'A'$ contains $6 \times 6 = 36$ sample points, $'B'$ contains $6$ sample points and $A\,\cap \,B$ contains only one sample point.
$\therefore P\left(A\,/\,B\right) = \frac{P\left(A\,\cap \,B\right)}{P\left(B\right)}$
$= \frac{\frac{1}{216}}{\frac{6}{216}} = \frac{1}{6}$