Probability — Class 12 Maths Solution

.: When a dice is thrown three times, then the sample space contains $6 \times 6 \times 6 = 216$

equally likely events. The sample space is $S = \{ (x,\;y,\;z):x,y,z \in \{ 1,2,3,4,5,6\} \}$

Let $E$ : 4 appears on the third toss
i.e., $E = \{ (x,y,4)$ : $x,y \in \{ 1,2,3,4,5,6\} \}$

and $F$ : $6$ and $5$ appears respectively on first two tosses

i.e., $F = \{ (6,5,1),\;(6,5,2),\;(6,5,3),\;(6,5,4),\;(6,5,5),\;(6,5,6)\}$

$\Rightarrow E \cap F = \{ (6,5,4)\}$

$P(E) = 36/216,P(F) = 6/216 = \cfrac{1}{{36}}$

and $P(E \cap F) = \cfrac{1}{{216}}$

Required probability, $P(E|F) = \cfrac{{P(E \cap F)}}{{P(F)}} = \cfrac{{1/216}}{{6/216}} = \cfrac{1}{6}$