91Ó°ÊÓ

• Conditional Probability- Probability of an event when one event already happened.

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

• Formula of probability $=\frac{\text{Number of favorable outcomes}}{\text{Total number of oulcomes}}.$

Calculation:

The tossing of two dice results in $36$outcomes.

Let ${}^{\text{'}}{E}^{\text{'}}$be the event that atleast one dice lands on $6$.

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

Let ${{}^{\text{'}}F}^{\text{'}}$be the event that both the numbers are different on the dice.

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

We have to find the probability of getting at least one dice lands on $6$and given that numbers on both the dice are different.