Question

In a single throw of pair of different dice what is the probability of getting a prime number on each dice .Atotal of 9or 11

Answer 1

As we throw two die we have 6^2 i.e 36 chances

So n(S) = 36

Let the event A be having prime on each dice

So chances are {(2,2), (2,3), (2,5), (3,2), (3,3), (3,5), (5,2), (5,3), (5,5)}

Hence n(A) = 9

Thus P(A) = 9/36 = 1/4

Let B be the event having the sum of 9 or 11

So chances will be like {(3,6), (6.3), (4.5), (5,4), (5,6), (6,5)}

So n(B) = 6

Thus P(B) = 6/36 = 1/6

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arumugamleylanowruoq

arumugamleylanowruoq

Answer:Let the event A be having prime on each dice

So chances are {(2,2), (2,3), (2,5), (3,2), (3,3), (3,5), (5,2), (5,3), (5,5)}

Hence n(A) = 9

Thus P(A)

= 9/36

= 1/4

Let B be the event having the sum of 9 or 11

So chances will be like

{(3,6), (6.3), (4.5), (5,4), (5,6), (6,5)}

So n(B) = 6

Thus P(B)

= 6/36

= 1/6