Question #11002

A committee of four is to be formed from among 5 economists, 6 engineers, 3 statisticians and 1 doctor . Find the probabilities that,

(i) Each profession is represented in the committee

(ii) It has a doctor and at least one economist

(iii) It has a doctor and at least one economist and others from remaining members

(i) Each profession is represented in the committee

(ii) It has a doctor and at least one economist

(iii) It has a doctor and at least one economist and others from remaining members

Expert's answer

A committee of four is to be formed from among 5 economists, 6 engineers, 3 statisticians and 1 doctor. Find the probabilities that,

(i) Each profession is represented in the committee

The total number of possible committees is С(5+6+3+1,4).

P = (С(5,1)+С(6,1)+С(3,1)+С(1,1))/С(5+6+3+1,4) = (5+6+3+1)/С(15,4) = 15/1365 ≈ 0.0110.

(ii) It has a doctor and at least one economist

There is C(5,1) = 5 ways to choose one economist and С(15-1-1,2) ways to choose others (excepting one economist and a doctor). So,

P = (С(5,1)+С(15-1-1,2))/С(5+6+3+1,4) = (5+C(13,2))/1365 = (5+78)/1365 ≈ 0.0608.

(iii) It has a doctor and at least one economist and others from remaining members

This item is similar to previous.

(i) Each profession is represented in the committee

The total number of possible committees is С(5+6+3+1,4).

P = (С(5,1)+С(6,1)+С(3,1)+С(1,1))/С(5+6+3+1,4) = (5+6+3+1)/С(15,4) = 15/1365 ≈ 0.0110.

(ii) It has a doctor and at least one economist

There is C(5,1) = 5 ways to choose one economist and С(15-1-1,2) ways to choose others (excepting one economist and a doctor). So,

P = (С(5,1)+С(15-1-1,2))/С(5+6+3+1,4) = (5+C(13,2))/1365 = (5+78)/1365 ≈ 0.0608.

(iii) It has a doctor and at least one economist and others from remaining members

This item is similar to previous.

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