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Exercise 1.12 . We roll a fair die repeatedly until we see the number four appe...

Sep 10, 2024

Solution by Steps

step 1

To find the probability of needing at most 3 rolls to get a four, we can use the geometric distribution. The probability of success (rolling a four) on each roll is

p = \frac{1}{6}

. The probability of needing at most

k

rolls is given by

P(X \leq k) = 1 - (1 - p)^k

. For

k = 3

step 2

Thus, we calculate:

P(X \leq 3) = 1 - \left(1 - \frac{1}{6}\right)^3 = 1 - \left(\frac{5}{6}\right)^3 = 1 - \frac{125}{216} = \frac{91}{216}

step 3

For part (b), we need to find the probability of needing an even number of rolls. The even rolls can be 2, 4, 6, etc. The probability of needing exactly

2n

rolls is given by

P(X = 2n) = (1 - p)^{2n-1} p

. We sum this for

n = 1, 2, \ldots

step 4

The probability of needing an even number of rolls is:

P(\text{even}) = \sum_{n=1}^{\infty} P(X = 2n) = \sum_{n=1}^{\infty} \left( \frac{5}{6} \right)^{2n-1} \left( \frac{1}{6} \right) = \frac{1/6}{1 - 5/6^2} = \frac{1/6}{1 - 25/36} = \frac{1/6}{11/36} = \frac{6}{11}

Answer

(a) The probability that we need at most 3 rolls is

\frac{91}{216}

. (b) The probability that we needed an even number of rolls is

\frac{6}{11}

Key Concept

The problem involves geometric distribution and calculating probabilities for discrete events.

Explanation

The solution uses the properties of the geometric distribution to find the probabilities of rolling a die until a specific outcome occurs.