Question

A new battery’s voltage may be acceptable (A) or unaccept- able (U). A certain flashlight requires two batteries, so bat- teries 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 is, P(Y 5 2)?

b) What is p(3)? [Hint: There are two different outcomes

that result in Y 5 3.]

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

selected? List the four outcomes for which Y 5 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)

e) instead of the random variable Y given in the textbook, define a Negative Binomial random variable and do this question. Also, state the mean and standard deviation for this negative binomial random variable and interpret it in the context of the question

Answer 1

From the given information we have

P(A) = 0.90, P(U) = 1 - 0.90 = 0.10

a)

Here we need to find the probability that is two trials we get two batteries that are acceptable.

The possible outcome is AA.

Since each battery is independent from other so

p(2)= P(AA)= P(A)P(A) =0.9 *0.9 = 0.81

b)

Here we need to find the probability that is three trials we get two batteries that are acceptable.

The possible outcomes are AUA and UAA.

Since each battery is independent from other so

p(3)= P(AUA)+P(UAA)= P(A)P(U)P(A)+P(U)P(A)P(A) =0.9 *0.10* 0.9 + 0.10*0.9 *0.9= 0.081+0.081 = 0.162

c)

Here we need to find the probability that is five trials we get two batteries that are acceptable.

The possible outcomes are AUUUA , UAUUA, UUAUA, UUUAA, .

Since each battery is independent from other so

p(5)= P(AUUUA )+P(UAUUA)+P(UUAUA)+P(UUUAA)= 0.9 *0.10*0.10*0.10* 0.9 + 0.10*0.9 *0.10*0.10*0.9 + 0.10 *0.10*0.90*0.10*0.90+0.10*0.10*0.10*0.90*0.90=0.00324

e)

Here Y has negative binomial distribution with parameters as follows;

p =0.90, r=2

The pdf of Y is:

The required probabilites are:

The mean is:

The variance is: