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 Y denote the number of batteries that must be tested. a. ?What is p(2), that isP ( Y = 2), ? b. ?What is p(3)? [Hint: There are two different outcomes that result inY = 3 .] c. ?To have Y=5, what must be true of the fifth battery selected? List the four outcomes for which Y= 5 and then determine p(5). d. ?Use the pattern in your answers for parts (a)–(c) to obtain a general formula for p(y).

Solution : Step 1 : Here a certain flashlight requires two batteries, so batteries will be independently selected and tested until two acceptable ones have been found. A battery’s voltage may be acceptable (A), or rejected. It is also given that 90% of all batteries are acceptable . and Y denote the number of batteries that must be tested. We have to find different probabilities P(2),P(5),P(3), and also want make a general formula for P(y). Step 2: a) We have to find the probability P(Y=2) , means probability that only two batteries must be tested. It is given that , The trials are identical, and each trial can result into A = { Acceptable voltage } B = { Unacceptable voltage } P(A)= P(batteries are acceptable) =0.9 , P(U) = P(batteries are not acceptable) = 1-0.9 =0.1 And here the trials are independent that means the the outcome of a trial does not influence the outcome of the other. So here the only possibility that only 2 of the batteries is to tested is both the selected batteries are acceptable. That means, P(Y=2)=P(A A) Since the trials are independent. P(y=2) =P(A)× P(A) = (0.9)2 = 0.81