Question

A computer center has three printers, A, B, and C, which print at different speeds.Programs are routed to the first available printer. The probability that a programis routed to printers A, B, and C are 0.6, 0.3, and 0.1, respectively. Occasionally aprinter will jam and destroy a printout. The probability that printers A, B, and C willjam are 0.01, 0.05 and 0.04, respectively. Your program is destroyed when a printerjams. What is the probability that printer A is involved? Printer B is involved?Printer C is involved?

Answer 1

Answer:

There is a 24% probability that printer A is involved.

There is a 60% probability that printer B is involved.

There is a 16% probability that printer C is involved.

Step-by-step explanation:

We have the following probabilities:

-A 60% probability that printer A is used.

-A 30% probability that printer B is used.

-A 10% probability that printer C is used.

-A 1% probability that printer A jams

-A 5% probability that printer B jams

-A 4% probability that printer C jams

This can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

$P = \frac{P(B).P(A/B)}{P(A)}$

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

Your program is destroyed when a printer jams. What is the probability that printer A is involved?

This question can be modeled as:

What is the probability that the printer A was used, knowing that the printer jammed. So:

P(B) is the probability that printer A is used.

So $P(B) = 0.6$

$P(A/B)$ is the probability that the printer jams when the printer A is used. So

$P(A/B) = 0.01$

P(A) is the probability that the printer jams. So

$P(A) = P_{1} + P_{2} + P_{3}$

$P_{1}$ is the probability that printer A is chosen and jams. So

$P_{1} = 0.6*0.01 = 0.006$

$P_{2}$ is the probability that printer B is chosen and jams. So

$P_{2} = 0.3*0.05 = 0.015$

$P_{3}$ is the probability that printer C is chosen and jams. So

$P_{2} = 0.1*0.04 = 0.004$

$P(A) = P_{1} + P_{2} + P_{3} = 0.006 + 0.015 + 0.004 = 0.025$

The probability that printer A is involved is:

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.6*0.01}{0.025} = 0.24$

There is a 24% probability that printer A is involved.

The probability that printer B is involved is:

P(A), that is the probability that the printer jams, is the same.

P(B) is now the probability that the printer B is chosen, so:

$P(B) = 0.3$

P(A/B) is the probability that the printer jams when printer B is chosen. So

$P(A/B) = 0.05$.

The probability that printer B is involved is:

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.3*0.05}{0.025} = 0.6$

There is a 60% probability that printer B is involved.

The probability that printer C is involved is:

P(A), that is the probability that the printer jams, is the same.

P(B) is now the probability that the printer C is chosen, so:

$P(B) = 0.1$

P(A/B) is the probability that the printer jams when printer C is chosen. So

$P(A/B) = 0.04$.

The probability that printer B is involved is:

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.1*0.04}{0.025} = 0.16$

There is a 16% probability that printer C is involved.