Question

A die is thrown three times,
$E:4$ appears on the third toss, $F:6$ and $5$ appears respectively on first two tosses.
Determine $36P$ where $P\left(E|F\right)$

Answer 1

If a die is thrown three times, then the number of elements in the sample space will be $6\times 6\times 6=216$

$E=\{(1,1,4),(1,2,4),\dots (1,6,4)(2,1,4),(2,2,4),\dots (2,6,4)(3,1,4),(3,2,4),\dots (3,6,4)(4,1,4),(4,2,4),\dots (4,6,4)(5,1,4),(5,2,4),\dots (5,6,4)(6,1,4),(6,2,4),\dots (6,6,4)\}$

$F=\{(6,5,1),(6,5,2),(6,5,3),(6,5,4),(6,5,5),(6,5,6)\}$

$\therefore E\cap F=\left\{(6,5,4)\right\}$

$P\left(F\right)=\frac{6}{216}$ and $P(E\cap F)=\frac{1}{216}$

$\therefore P\left(E|F\right)=\frac{P(E\cap F)}{P\left(F\right)}=\frac{\frac{1}{216}}{\frac{6}{216}}=\frac{1}{6}=0.16$