Answer:
The probability that you test exactly 4 batteries is 0.0243.
Step-by-step explanation:
We are given that 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 90% of all batteries have acceptable voltages.
Let the probability that batteries have acceptable voltages = P(A) = 0.90
So, the probability that batteries have unacceptable voltages = P(U) = 1 - P(A) = 1 - 0.90 = 0.10
Now, the probability that you test exactly 4 batteries is given by the three cases. Firstly, note that batteries will be tested until two acceptable ones have been found.
So, the cases are = P(AUUA) + P(UAUA) + P(UUAA)
This means that we have tested 4 batteries until we get two acceptable batteries.
So, required probability = (0.90 
0.10 0.10 0.90) + (0.10 0.90 0.10 0.90) + (0.10 0.10 0.90 0.90)
= 0.0081 + 0.0081 + 0.0081 = <u>0.0243</u>
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Hence, the probability that you test exactly 4 batteries is 0.0243.
