A pair of fair dice is rolled together till a sum of either 5 or 7 is obtained. If p denoted the probability that 7 comes before 5, find 15p?

Solution:

Assume A as the event that a sum of 7 occurs

B as the event that a sum of 5 occurs

C as the event that neither a sum of 5 nor a sum of 7 occurs

We know that

P (A) = 6/36 = 1/6

P (B) = 4/36 = 1/9

P (C) = 26/36 = 13/18

Here

p = P (A or (C ∩ A) or (C ∩ C ∩ A) or ….. )

= P (A) + P(C) P(A) + P(C)2 P(A) + ….

= P(A)/ [1 - P(C)]

Substituting the values

= 1/6/ [1 - 13/18]

= 3/5

15p = 3/5 × 15 = 9

Therefore, 15p is 9.

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A pair of fair dice is rolled together till a sum of either 5 or 7 is obtained. If p denoted the probability that 7 comes before 5, find 15p?

Summary:

A pair of fair dice is rolled together till a sum of either 5 or 7 is obtained. If p denotes the probability that 7 comes before 5, 15p is 9.