find the probability

Question

A committee of 4 people is to be appointed from 5 officers of the production department, 4 officers of the purchase department, two officers of the sales department and 1 charted accounted. Find the probability of forming the committee where we must be one from each category.

Answer 1

Label the 4 departments A (production), B (purchase), C (sales), D (accountancy), then the order of selection corresponds to ABCD, ABDC, ACBD, etc. As each officer is selected, the personnel pool reduces by one, so the probabilities fluctuate.

Here are the 24 ways to order the selection with their corresponding probabilities:

ABCD, ABDC, ACBD, ACDB, ADBC, ADCB=

(5/12)(4/11)(2/10)(1/9), (5/12)(4/11)(1/10)(2/9), (5/12)(2/11)(4/10)(1/9),

(5/12)(2/11)(1/10)(4/9), (5/12)(1/11)(4/10)(2/9), (5/12)(1/11)(2/10)(4/9);

BACD, BADC, BCAD, BCDA, BDAC, BDCA=

(4/12)(5/11)(2/10)(1/9), (4/12)(5/11)(1/10)(2/9), (4/12)(2/11)(5/10)(1/9),

(4/12)(2/11)(1/10)(5/9), (4/12)(1/11)(5/10)(2/9), (4/12)(1/11)(2/10)(5/9);

CABD, CADB, CBAD, CBDA, CDAB, CDBA=

(2/12)(5/11)(4/10)(1/9), (2/12)(5/11)(1/10)(4/9), (2/12)(4/11)(5/10)(1/9),

(2/12)(4/11)(1/10)(5/9), (2/12)(1/11)(5/10)(4/9), (2/12)(1/11)(4/10)(5/9);

DABC, DACB, DBAC, DBCA, DCAB, DCBA=

(1/12)(5/11)(4/10)(2/9), (1/12)(5/11)(2/10)(4/9), (1/12)(4/11)(5/10)(2/9),

(1/12)(4/11)(2/10)(5/9), (1/12)(2/11)(5/10)(4/9), (1/12)(2/11)(4/10)(5/9).

Each probability has the same value, which can be reduced to:

(1/3)(1/11)(1/9)=1/297. Total sum=24/297=8/99=8.08% approx.