In a single throw of two dice, find

Solution:

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 $(4,4),(4,5),(4,6),(5,4),(5,5),(5,6),(6,4),(6,5),(6,6)$

Total no.of outcomes are 36 and desired outcomes are 9

Therefore, probability of getting number greater than 3 on each die

$=\frac{9}{36}$

$=\frac{1}{4}$

Conclusion: Probability of getting a number greater than 3 on each die, when two dice are rolled is $\frac{1}{4}$