Question

A homeowner finds that there is a 0.15 probability that a flashlight does not work when turned on. If she has three flashlights, find the probability that at least one of them works when there is a power failure. Find the probability that the second flashlight works given that the first flashlight works.

Answer 1

Answer:0.9966

Step-by-step explanation:

Given

Probability that flash light does not  work is $P_o=0.15$

If owner has 3 three flashlights then

Probability that atleast one of them works $=1-P(\text{none of them works})$

Probability that flashlight will work $=1-P_o=1-0.15$

$=0.85$

Required Probability$=1-0.15\times 0.15\times 0.15$

$=1-0.003375$

$=0.9966$

Now, Probability that second works given that first works is given by

$P=P(\text{First works})\times P(\text{Second works})$

$P=0.85\times 0.85$

$P=0.7225$