Hiiii pls help me if u can tysm

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: