Question

In a photocopy center, there are two small photocopiers, two medium photocopiers, and one big photocopier. The probability that a small one fails and requires repairs is 0.1, a medium one fails and requires repairs is 0.08, and the probability that the big photocopier fails and requires repairs is 0.05. Assume that all five copiers operate independently.

a. What is the probability that all the photocopiers fail and require repairs?
b. What is the probability that none of the photocopiers fails and requires repairs?
c. What is the probability that one of the photocopiers fails and requires repairs?
d. What is the probability that one of the small photocopiers fails and requires repairs?
e. What is the probability that at least one of the five copiers fails and requires repairs?

Answer 1

a. The probability that all the photocopiers fail and require repairs is 0.000032. b. The probability that none of the photocopiers fails and requires repairs is 0.36. c. The probability that one of the photocopiers fails and requires repairs is 0.2460256. d. The probability that one of the small photocopiers fails and requires repairs is 0.1. e. The probability that at least one of the five copiers fails and requires repairs is 0.64.

Step-by-step explanation:

a. What is the probability that all the photocopiers fail and require repairs?

To find the probability that all the photocopiers fail and require repairs, we need to multiply the individual probabilities of each copier failing. So, the probability that all the photocopiers fail is:

0.1 * 0.1 * 0.08 * 0.08 * 0.05 = 0.000032

b. What is the probability that none of the photocopiers fails and requires repairs?

To find the probability that none of the photocopiers fails, we need to find the complement of the probability that at least one copier fails. So, the probability that none of the photocopiers fails is:

1 - (0.9 * 0.9 * 0.92 * 0.92 * 0.95) = 0.36

c. What is the probability that one of the photocopiers fails and requires repairs?

To find the probability that one of the photocopiers fails, we need to find the probability that exactly one copier fails. This can happen in multiple ways:

P(Small copier 1 fails) * P(no other copier fails) = 0.1 * 0.9 * 0.92 * 0.92 * 0.95 = 0.078888

P(Small copier 2 fails) * P(no other copier fails) = 0.1 * 0.1 * 0.92 * 0.92 * 0.95 = 0.008416

P(Medium copier 1 fails) * P(no other copier fails) = 0.9 * 0.08 * 0.92 * 0.92 * 0.95 = 0.0611936

P(Medium copier 2 fails) * P(no other copier fails) = 0.9 * 0.92 * 0.08 * 0.92 * 0.95 = 0.0611936

P(Big copier fails) * P(no other copier fails) = 0.9 * 0.92 * 0.92 * 0.92 * 0.05 = 0.0363336

Adding up all these probabilities, we get the probability that exactly one copier fails is: 0.078888 + 0.008416 + 0.0611936 + 0.0611936 + 0.0363336 = 0.2460256

d. What is the probability that one of the small photocopiers fails and requires repairs?

To find the probability that one of the small photocopiers fails, we need to find the probability that exactly one small copier fails. This can happen in two ways:

P(Small copier 1 fails) * P(no other small copier fails) = 0.1 * 0.9 = 0.09

P(Small copier 2 fails) * P(no other small copier fails) = 0.1 * 0.1 = 0.01

Adding up these probabilities, we get the probability that exactly one small copier fails is: 0.09 + 0.01 = 0.1

e. What is the probability that at least one of the five copiers fails and requires repairs?

This is the complement of the probability that none of the copiers fail. So, the probability that at least one copier fails is: 1 - (0.9 * 0.9 * 0.92 * 0.92 * 0.95) = 0.64