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A fair die is rolled repeatedly until a six appears for the first time. The probability that the number of rolls needed is even equals p/q in lowest…

Numerical answer

A fair die is rolled repeatedly until a six appears for the first time. The probability that the number of rolls needed is even equals

\dfrac pq

in lowest terms. The value of

p+q

Show answer and solution

Answer: 16

The first six occurs on roll

k

with probability

\left(\dfrac56\right)^{k-1}\dfrac16

. The even values of

k

are

2,4,6,\ldots

, so the required probability is

\dfrac16\left[\dfrac56+\left(\dfrac56\right)^3+\left(\dfrac56\right)^5+\cdots\right]

, an infinite geometric series with first term

\dfrac56

and ratio

\dfrac{25}{36}

. Its sum is

\dfrac{5/6}{11/36}=\dfrac{30}{11}

, so the probability is

\dfrac16\cdot\dfrac{30}{11}=\dfrac5{11}

. Hence

p=5

q=11

and

p+q=16

. (Check by symmetry: if

P

is the even probability then

P=\tfrac56(1-P)

gives

P=\tfrac5{11}

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