In a single throw of two dice, find

We know that,

Probability of occurrence of an event

$=\frac{\text { Total no.of Desired outcomes }}{\text { Total no.of outcomes }}$

Total outcomes are $(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)$, $(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)$, $(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)$, $(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)$, $(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)$, $(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)$

Desired outcomes are $(3,6),(4,5),(5,4),(6,3),(6,5),(5,6)$ Total no. of outcomes are 36 and desired outcomes are 6

Therefore, probability of getting total equal to 9 or $11=\frac{6}{36}$

$=\frac{1}{6}$

Conclusion: Probability of getting total equal to 9 or 11 , when two dice are rolled is $\frac{1}{6}$