geometric distribution

Question

The probability of obtaining a defective 10-year old widget is 64.1%. For our purposes, the random variable will be the number of items that must be tested before finding the first defective 10-year old widget. Thus, this procedure yields a geometric distribution.

Find the probability distribution.

(Report answers accurate to 4 decimal places.) The Second column is P(X = k)

k
1	Correct
2
3
4
5
6 or greater

Answer 1

We are pulling one widget at a time and determining whether it is defective or not. There are only two potential outcomes. Let D mean defective and N mean not defective. We will pull one at a time and stop once we find a defective piece.

So Pr(D) = .641, and Pr(N) = 1 - .641 = .359

So here are the possibilities.

D k=1 Pr(1) =.359

ND k=2 Pr(2) = (.641)( .359)

NND k=3 Pr(3) = (.641)(.359)^2

NNND k=4 Pr(4) = (.641)(.359)^3

NNNND k=5 Pr(5) = (.641)(.359)^4

NNNNND k=6 Pr(6) = (.641)(.359)^5

NNN.........D k=x Pr(x) = (.641)(.359)^(x-1)

Answer 2

Jessica, for this problem and the next one, you have the formula available, P(k) = p*((1-p)^(k-1))

(I put in the * and extra parentheses in because, while the formula is correct without it, you would, in the case of the *, get the wrong answer plugging it into a computer or calculator, and, in the case of the extra parentheses, avoid the chance of entry errors; this also makes the formula easier to understand).

The tricky one may be the 6 or greater. There are 2 ways to solve this. The long way would be 1 - P(1) - P(2)-P(3)-P(4)-P(5). The shorter way is to understand that requiring 6 or more widgets means that all of the first five were not defective. That would be (1 - .641)^5