Answer:
(a) 0.62
(b) 0.2244
(c) 0.5097
Step-by-step explanation:
(a) The probability of a failure in a shift is : 0.38
The probability of no failure during a shift will thus be: 1 - 0.38 = 0.62
(b) Number of shifts in a 48 hour period = 48 / 8 = 6 shifts
Number of ways 4 failures could occur in 6 shifts (permutations with similar objects) = 6! / (4!) = 24
Probability of 4 shifts with failures and 2 shifts without:
0.38 * 0.38 * 0.38 * 0.38 * 0.62 * 0.62 = 0.008
This means there is a total probability of 0.008 * 24 = 0.192 probability of 4 failures occurring in 6 shifts.
Similarly for 5 and 6 failures we calculate as follows:
Number of ways 5 failures can occur = 6! / 5! = 6
Probability of 5 failures = 0.38^5 * 0.62 = 0.0049
Total probability of 5 failures = 6 * 0.0049 = 0.0294
Number of ways 6 failures can occur = 1
Probability of 6 failures = 0.38^6 = 0.003
Adding probabilities of 4, 5 and 6 failures together we get the total probability for (b) = 0.192 + 0.0294 + 0.003 = <u>0.2244</u>
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(c) Since there is at least one failure in a 24 hour period, the number of failures in a 48 hour period must be 2.
Given this fact, let's calculate the probability of 3 failures. Here we will only consider the 4 shifts in which failure is not given already.
Probability of exactly 1 failure in the 4 shifts = 0.38 * 0.62^3 = 0.0905
We must also multiply this probability with the number of ways this can occur.
Number of ways this can occur : 4! / 3! = 4
So probability of 3 failures = 0.0905 * 4 = 0.362
The probability of only two failures = 0.62 ^ 4 = 0.1477
Total probability of no more than 3 failures given there are already two failures = 0.362 + 0.1477 = <u>0.5097</u>
