Question

If $Y$ has a geometric distribution with success probability 0.3,what is the largest value of $y$,such that $P9Y>y)>0.1$ ?

Answer 1

Given $p=0.3$

Then $q=1-p=1-0.3=0.7$

By using the result $P(y>y_0)=q^{y_0}$, we have

$$P(y>y_0)=0.7^{y_0}$$

If $P(y>y_0)>0.1,$ then

$$0.7^{y_0}>0.1$$

Let $0.7^{y_0}=0.1$

$$\ln{0.7^{y_0}}=\ln{0.1}$$

$$y_0\ln{0.7}=\ln{0.1}$$

$$y_0=\dfrac{\ln 0.1}{\ln 0.7}\approx6.4557$$

$$0.7^{6}=0.117649>0.1$$

$$0.7^{7}=0.0823543<0.1$$

The largest value of $y$ is 6.