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

The probability that at least one flashlight works when there is a power failure is 0.996625. The probability that the second flashlight works given that the first flashlight works is 0.85.

Step-by-step explanation:

To find the probability that at least one flashlight works when there is a power failure, we can find the probability that all three flashlights do not work and subtract it from 1. Since the probability that a flashlight does not work is 0.15, the probability that all three flashlights do not work is 0.15 * 0.15 * 0.15 = 0.003375. Therefore, the probability that at least one flashlight works is 1 - 0.003375 = 0.996625.

To find the probability that the second flashlight works given that the first flashlight works, we can use the concept of conditional probability. Since one flashlight is already working, there are only two remaining flashlights. The probability that the second flashlight works is therefore 1 - the probability that it does not work. Since the probability that a flashlight does not work is 0.15, the probability that the second flashlight does work given that the first flashlight works is 1 - 0.15 = 0.85.

Answer 2

A) 0.386

B).0.325

Step-by-step explanation:

probability p will not work = 0.15

Q probability w5ill work = 0.85

Number of flashlight = 3

Probabilty of at least one=

1- probability of none

But probability of none= 3C0(p)^0(q)^3

= 1*(0.15)^0(0.85)^3

= 1*1*(0.614)

= 0.614

Probability of at least one=1-0.614

= 0.386

Probability of two flashlight work

= 3C1(0.15)^1(0.85)^2

= 3*(0.15)*(0.7225)

= 0.325