Answer:
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: