A new battery's voltage may be acceptable (A) or unacceptable (U). A certain flashlight requires two batteries, so batteries will be independently selected and tested until two acceptable ones have been found. Suppose that 92% of all batteries have acceptable voltages. Let Y denote the number of batteries that must be tested.

a) What is p(2), that is P(y = 2)?

b) What is p(3)?

c) To have Y = 5, what must be true of the fifth battery selected?

i. The fifth battery must be an A.

ii. The fifth battery must be a U.

Given:

92% batteries have acceptable voltages

Let Y denotes the number of batteries that must be tested.

P(acceptable)=P(A)=0.92P(acceptable)=P(A)=0.92

P(unacceptable)=1−P(A)=0.08P(unacceptable)=1−P(A)=0.08\

(a)

P(y=2)=P(AA)=0.92×0.92=0.8464P(y=2)=P(AA)=0.92×0.92=0.8464

(b)

The favorable cases to y=3 are AUA, UAA

P(y=3)=P(AUA)+P(UAA)=(0.92×0.08×0.92)+(0.08×0.92×0.92)=0.1354P(y=3)=P(AUA)+P(UAA)=(0.92×0.08×0.92)+(0.08×0.92×0.92)=0.1354

(c)

In order to have y=5 the 5th battery must be second acceptable battery

The favorable outcomes are AUUUA, UUUAA, UAUUA and UUAUA

P(y=5)=P(AUUUA)+P(UUUAA)+P(UAUUA)+P(UUAUA)=0.92×0.08×0.08×0.08×0.92+0.08×0.92×0.08×0.08×0.92+0.08×0.92×0.08×0.08×0.92+0.08×0.08×0.92×0.08×0.92=0.0016

Step-by-step explanation:

4.08%

step-by-step explanation: