Question

A machine has four components, A, B, C, and D, set up in such a manner that all four parts must work for the machine to work properly. Assume the probability of one part working does not depend on the functionality of any of the other parts. Also, assume that the probabilities of the individual parts working are P(A) = P(B) = 0.93, P(C) = 0.95, and P(D) = 0.92. Find the probability that the machine works properly.

Answer 1

The probability that the machine works properly is found by multiplying the probabilities of all four components working: P(A) * P(B) * P(C) * P(D) = 0.93 * 0.93 * 0.95 * 0.92 = 0.7513, or 75.13%.

Step-by-step explanation:

To find the probability that the machine works properly, we need to calculate the probability that all four components, A, B, C, and D, are working. Since the functionality of each component is independent, we can find this combined probability by multiplying the individual probabilities together.

The probability of A working is P(A) = 0.93, B working is P(B) = 0.93, C working is P(C) = 0.95, and D working is P(D) = 0.92. So the probability of the machine working is:

P(Machine works) = P(A) * P(B) * P(C) * P(D) = 0.93 * 0.93 * 0.95 * 0.92 = 0.7513

Therefore, the probability that the machine works properly is 0.7513, which is 75.13%.

Answer 2

0.756

Step-by-step explanation:

It is given that a machine has four components, A, B, C, and D.

`P(A)=P(B)=0.93, P(C)=0.95,P(D)=0.92`

If these components set up in such a manner that all four parts must work for the machine to work properly.

We need to find the probability that the machine works properly. It means we have to find the value of
`P(A\cap B\cap C\cap D)`.

If two events X and Y are independent, then

`P(X\cap Y)=P(X)* P(Y)`

Assume the probability of one part working does not depend on the functionality of any of the other parts.

`P(A\cap B\cap C\cap D)=P(A)* P(B)* P(C)* P(D)`

Substitute the given values.

`P(A\cap B\cap C\cap D)=0.93* 0.93* 0.95* 0.92`

`P(A\cap B\cap C\cap D)=0.7559226`

`P(A\cap B\cap C\cap D)\approx 0.756`

Therefore, the probability that the machine works properly is 0.756.