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.

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

i'm pretty sure it's 477,585,296, but perhaps my math is incorrect.