Question

A committee of four is to be formed from among 5 economists, 4 engineers, 2 statisticians and 1 doctor . Find the probabilities that,
(i) Each profession is represented in the committee
(ii) Has atleast one economists
(iii) Has a doctor as a member and three others
(Iv) Has atleast one engineer but no economists

Answer 1

Answer on Question #69778 – Math – Statistics and Probability

Question

A committee of four is to be formed from among 5 economists, 4 engineers, 2 statisticians and 1 doctor. Find the probabilities that

(i) Each profession is represented in the committee

(ii) Has at least one economists

(iii) Has a doctor as a member and three others

(iv) Has at least one engineer but no economists

Solution

$$5 + 4 + 2 + 1 = 12$$

The total number of selections is

$$\binom{12}{4} = \frac{12!}{(12 - 4)!4!} = \frac{12(11)(10)(9)}{1(2)(3)(4)} = 495$$

(i) Each profession is represented in the committee

The following table shows the number of selections from each group:

Thus, a committee consisting of one profession from each field can be selected in

$$\binom{5}{1} \cdot \binom{4}{1} \cdot \binom{2}{1} \cdot \binom{1}{1} = 5(4)(2)(1) = 40$$

The probability that each profession is represented in the committee

$$P(A) = \frac{40}{495} = \frac{8}{99}$$

(ii) Has at least one economist

Find the probability that there is no economist in the committee

$$4 + 2 + 1 = 7$$

The suitable number of selections is

$$\binom{7}{4} = \frac{7!}{(7 - 4)!4!} = \frac{(7)(6)(5)}{1(2)(3)} = 35$$

The probability that there is no economist in the committee

$$P(\overline{B}) = \frac{35}{495} = \frac{7}{99}$$

The probability that there is at least one economist in the committee

$$P(B) = 1 - P(\overline{B}) = 1 - \frac{7}{99} = \frac{92}{99}$$

(iii) Has a doctor as a member and three others

The following table shows the number of selections from each group:

$$5 + 4 + 2 = 11$$

Thus, a committee consisting of one profession from each field can be selected in

$$\binom{11}{3} \cdot \binom{1}{1} = \frac{11!}{(11 - 3)! \cdot 3!} (1) = \frac{(11)(10)(9)}{1(2)(3)} = 165$$

The probability that the committee has a doctor as a member and three others

$$P(C) = \frac{165}{495} = \frac{1}{3}$$

(iv) Has at least one engineer but no economists

The following table shows the number of selections from each group:

Thus, a committee having at least one engineer but no economists can be selected in

$$\begin{array}{l} \binom{4}{1} \cdot \binom{2}{2} \cdot \binom{1}{1} + \binom{4}{2} \cdot \binom{2}{2} + \binom{4}{2} \cdot \binom{2}{1} \cdot \binom{1}{1} + \binom{4}{3} \cdot \binom{2}{1} + \binom{4}{3} \cdot \binom{1}{1} + \binom{4}{4} = \\ = 4 + \frac{4!}{(4 - 2)! (2)!} + \frac{4!}{(4 - 2)! (2)!} (2) + \frac{4!}{(4 - 3)! (3)!} (2) + \frac{4!}{(4 - 3)! (3)!} + 1 = \\ = 4 + 6 + 12 + 8 + 4 + 1 = 35 \\ \end{array}$$

The probability that the committee has at least one engineer but no economists

$$P(D) = \frac{35}{495} = \frac{7}{99}$$

Answer: i) $\frac{8}{99}$ ; ii) $\frac{92}{99}$ ; iii) $\frac{1}{3}$ ; iv) $\frac{7}{99}$.

Answer provided by https://www.AssignmentExpert.com