91Ó°ÊÓ

Given : Tossing of two dice and numbers on both the dice are different.

Conditional Probability: Conditional Probability is the probability of an event occurring after another has occurred.

$\mathrm{P}(\mathrm{E}/\mathrm{F})=\frac{\mathrm{P}(\mathrm{E}âˆ\copyright \mathrm{F})}{\mathrm{P}\left(\mathrm{F}\right)}.$

Probability of an event $=\frac{Numberoffavorableoutcomes}{Totalnumberofoutcomes}$

The tossing of two dice result in 36 outcomes.

Let ' $\mathrm{E}$' be the event that at least one dice lands on $6$.

Sample space for events' $E$' are $(1,6),(2,6),(3,6),(4,6),(5,6)\&(6,6)$

Let $F$be the event that both numbers on the dice are different.

Sample spaces for events are

$\begin{array}{r}(1,2),(1,3),(1,4),(1,5),(1,6)\\ (2,1),(2,3),(2,4),(2,5),(2,6)\\ (3,1),(3,2),(3,4),(3,5),(3,6)\\ (4,1),(4,2),(4,3),(4,5),(4,6)\\ (5,1),(5,2),(5,3),(5,4),(5,6)\\ (6,1),(6,2),(6,3),(6,4),(6,5)\end{array}$

$\mathrm{P}(\mathrm{E}/\mathrm{F})=\mathrm{P}($Event $\mathrm{E}$when the event $\mathrm{F}$is given )localid="1648221210413" $=\frac{\mathrm{P}\left(\mathrm{F}n\mathrm{F}\right)}{\mathrm{P}\left(F\right)}.$

$\mathrm{E}âˆ\copyright \mathrm{F}=$common of event $\mathrm{E}$and$\mathrm{F}=(1,6),(2,6),(3,6),(4,6)\mathrm{\&}(5,6)$. So

$\mathrm{P}(\mathrm{E}âˆ\copyright \mathrm{F})=\frac{5}{36},\mathrm{P}\left(\mathrm{F}\right)=\frac{30}{36}.$

Therefore,

localid="1648563788809" $\begin{array}{c}\mathrm{P}(\mathrm{E}/\mathrm{F})=\frac{\mathrm{P}(\mathrm{F}âˆ\copyright \mathrm{F})}{\mathrm{P}\left(\mathrm{F}\right)}\\ =\frac{\frac{5}{36}}{\frac{30}{36}}\\ =\frac{5}{36}×\frac{36}{30}\\ =\frac{1}{6}\end{array}$