Question

A weapons manufacturer uses a liquid propellant to produce gun car- tridges. During the manufacturing process, the propellant can get mixed with another liquid to produce a contaminated cartridge. A University of South Florida statistician, hired by the company to inves- tigate the level of contamination in the stored cartridges, found that 23% of cartridges in a particular lot were contaminated. Suppose you randomly sample (without replacement) gun cartridges from this lot until you nd a contaminated one. Let x be the number of cartridges sampled until a contaminated one is found. It is known that the probability distribution for x is given by the formula:
P(x) = (.23)(.77)x-1, x = 1,2,3,....
A. Find p1.
B. Find p5.
C, Find P{xk's are greater than or equal to 2}.

Answer 1

Answer:

$P(1) = 0.23$

$P(5) = 0.0809$

$P(x\ge 2) = 0.471$

Step-by-step explanation:

Given

$P(x) = 0.23 * 0.77^{x-1}$

Solving (a): P(1)

This gives:

$P(x) = 0.23 * 0.77^{x-1}$

$P(1) = 0.23 * 0.77^{1-1}$

$P(1) = 0.23 * 0.77^0$

$P(1) = 0.23 * 1$

$P(1) = 0.23$

Solving (b): P(5)

This gives:

$P(x) = 0.23 * 0.77^{x-1}$

$P(5) = 0.23 * 0.77^{5-1}$

$P(5) = 0.23 * 0.77^{4}$

$P(5) = 0.23 * 0.35153041$

$P(5) = 0.0809$

Solving (c): $P(x \ge 2)$

To do this, we make use of complement rule

$P(x \ge 2) = 1 - P(x < 2)$

$P(x < 2)$ is calculated as:

$P(x < 2) = P(0) +P(1)$

Calculate P(0)

$P(x) = 0.23 * 0.77^{x-1}$

$P(0) = 0.23 * 0.77^{0-1}$

$P(0) = 0.23 * 0.77^{-1}$

$P(0) = 0.299$

In (a):

$P(1) = 0.23$

So:

$P(x < 2) = P(0) +P(1)$

$P(x < 2) = 0.299 + 0.23$

$P(x < 2) = 0.529$

So:

$P(x \ge 2) = 1 - P(x < 2)$

$P(x\ge 2) = 1 - 0.529$

$P(x\ge 2) = 0.471$