In a single throw of two dice, find P (a total greater than 8)
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 $(3,6),(4,5),(4,6),(5,4),(5,5),(5,6),(6,3),(6,4),(6,5),(6,6)$
Total no. of outcomes are 36 and desired outcomes are 10
Therefore, probability of getting total greater than $8=\frac{10}{36}$
$=\frac{5}{18}$
Conclusion: Probability of getting total greater than 8, when two dice are rolled is $\frac{5}{18}$
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