A committee will be formed with 2 managers and 4 engineers selected randomly wit
ID: 3068011 • Letter: A
Question
A committee will be formed with 2 managers and 4 engineers selected randomly without replacement from 11 managers and 19 engineers (a) What is the probability that the specific engineer Jane and the specific manager Mary are on the committee? Round your answer to three decimal places (e.g. 98.765) p l.21 (b) Each of the engineers differ in years of experience. What is the probability that the most experienced and least experienced engineers are on the committee? Round your answer to four decimal places (e.g. 98.7654)Explanation / Answer
a) There are 11 managers. So the probability of a specified manager included in the sample is = 2 / 11
Because we want to select 2 managers out of 11 managers.
Similarly the probability of a specified engineer included in the sample is = 4 / 19
Because we want to select 4 engineers out of 11 engineers .
So the required probability = (2/11)*(4/19) = 0.038
b) Suppose the most experienced and the least experienced engineer will be selected, then we need two more engineer out of remaining 17 engineers.
Therefore possible ways = 1* 17C2 = 136
And the total ways of selecting 4 engineers from the 19 engineers = 19C4 = 3876
Therefore required probability = 136/3876 = 0.0351
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