A machine has 7 identical components which function independently. The probability that a component will fail is .2. The machine will stop working if more than three components fail. Find the probability that the machine will be working?
The machine will be working if
1. no part fails ---> (.8)^7
2. 1 part fails ----> C(7,1)(.2)(.8)^6
3. 2 parts fail ---> C(7,2)(.2^2)(.8^5)
4. 3 parts fail ---> C(7,3)(.2^3)(.8^4)
add them up
.966
To find the probability that the machine will be working, we need to consider the cases where 0, 1, 2, or 3 components fail, and then sum up the probabilities of these cases.
Let's break it down:
Case 0: All 7 components are working.
Case 1: Exactly 1 component fails, and the other 6 are working.
Case 2: Exactly 2 components fail, and the other 5 are working.
Case 3: Exactly 3 components fail, and the other 4 are working.
Since the components function independently, we can use the binomial probability formula to calculate the probability for each case:
P(k successes out of n trials) = (n choose k) * p^k * (1-p)^(n-k)
In this case, n = 7 (number of components), p = 0.2 (probability of failure), and k = 0, 1, 2, or 3.
Now let's calculate each case:
Case 0: P(0 components fail) = (7 choose 0) * 0.2^0 * (1-0.2)^(7-0) = 1 * 1 * 0.8^7 = 0.2097
Case 1: P(1 component fails) = (7 choose 1) * 0.2^1 * (1-0.2)^(7-1) = 7 * 0.2 * 0.8^6 = 0.3359
Case 2: P(2 components fail) = (7 choose 2) * 0.2^2 * (1-0.2)^(7-2) = 21 * 0.04 * 0.8^5 = 0.3020
Case 3: P(3 components fail) = (7 choose 3) * 0.2^3 * (1-0.2)^(7-3) = 35 * 0.008 * 0.8^4 = 0.0576
Now we can sum up the probabilities of these cases to find the probability that the machine will be working:
P(machine working) = P(0 components fail) + P(1 component fails) + P(2 components fail) + P(3 components fail)
= 0.2097 + 0.3359 + 0.3020 + 0.0576
≈ 0.9052
Therefore, the probability that the machine will be working is approximately 0.9052 or 90.52%.