Answer to Question #137141 in Statistics and Probability for Somila Ngangana

Question

A box of 10 light bulbs, 2 of which are faulty. A bulb is selected at random and put into a socket. If it works it is left there, if it is faulty it is discarded and a new bulb is selected. The process I continued until a bulb which works is found. Determine the probability distribution of X, the number of bulbs tried.

Answer 1

solution

Let the event that a bulb works be a success

$P(success) = \frac{8}{10}=0.8$

$P(failure) = \frac{2}{10}=0.2$

X follows a geometric distribution i.e. the number of trials before the first success

$f(x)=(1-p)^{x-1}p$ $f(x)=(0.2)^{x-1}*0.8$

The number of possible trials before the first success is

$X: \{1,\ 2,\ 3\}$

$f(X): \ 0.8, \ 0.16,\ 0.032$