Question

(4). Suppose Komla throws a die repeatedly until he gets a six. What is the probability
that he needs to throw more than 10 times to get a six, to 3 decimal places?

Answer 1

Geometric distribution

If repeated independent trials can result in a success with probability $p$ and a failure with probability $q=1-p,$ then the probability distribution of the random variable $X,$ the number of the trial on which the first success occurs, is

$$g(x;p)=p(1-p)^{x-1}, x=1,2,3,...$$

Given $p=\dfrac{1}{6}$

$$P(X=1)={1 \over 6}(1-{1 \over 6})^0={1 \over 6}$$

$$P(X=2)={1 \over 6}(1-{1 \over 6})^1={1 \over 6}({5 \over 6})^1$$

$$...$$

$$P(X=10)={1 \over 6}(1-{1 \over 6})^{9}={1 \over 6}({5 \over 6})^{9}$$

$$P(X\leq10)=\displaystyle\sum_{i=1}^{10}P(X=x_i)={1 \over 6}\displaystyle\sum_{i=1}^{10}({5 \over 6})^{i-1}=$$

$$={1 \over 6}\cdot\dfrac{1-({5 \over 6})^{10}}{1-{5\over 6}}=1-({5 \over 6})^{10}$$

$$P(X>10)=1-P(X\leq10)=({5 \over 6})^{10}\approx0.162$$