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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}.

Respuesta :

MrRoyal

Answer:

[tex]P(1) = 0.23[/tex]

[tex]P(5) = 0.0809[/tex]

[tex]P(x\ge 2) = 0.471[/tex]

Step-by-step explanation:

Given

[tex]P(x) = 0.23 * 0.77^{x-1}[/tex]

Solving (a): P(1)

This gives:

[tex]P(x) = 0.23 * 0.77^{x-1}[/tex]

[tex]P(1) = 0.23 * 0.77^{1-1}[/tex]

[tex]P(1) = 0.23 * 0.77^0[/tex]

[tex]P(1) = 0.23 * 1[/tex]

[tex]P(1) = 0.23[/tex]

Solving (b): P(5)

This gives:

[tex]P(x) = 0.23 * 0.77^{x-1}[/tex]

[tex]P(5) = 0.23 * 0.77^{5-1}[/tex]

[tex]P(5) = 0.23 * 0.77^{4}[/tex]

[tex]P(5) = 0.23 * 0.35153041[/tex]

[tex]P(5) = 0.0809[/tex]

Solving (c): [tex]P(x \ge 2)[/tex]

To do this, we make use of complement rule

[tex]P(x \ge 2) = 1 - P(x < 2)[/tex]

[tex]P(x < 2)[/tex] is calculated as:

[tex]P(x < 2) = P(0) +P(1)[/tex]

Calculate P(0)

[tex]P(x) = 0.23 * 0.77^{x-1}[/tex]

[tex]P(0) = 0.23 * 0.77^{0-1}[/tex]

[tex]P(0) = 0.23 * 0.77^{-1}[/tex]

[tex]P(0) = 0.299[/tex]

In (a):

[tex]P(1) = 0.23[/tex]

So:

[tex]P(x < 2) = P(0) +P(1)[/tex]

[tex]P(x < 2) = 0.299 + 0.23[/tex]

[tex]P(x < 2) = 0.529[/tex]

So:

[tex]P(x \ge 2) = 1 - P(x < 2)[/tex]

[tex]P(x\ge 2) = 1 - 0.529[/tex]

[tex]P(x\ge 2) = 0.471[/tex]

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