Answer to Question #211479 in Statistics and Probability for Farah

Question

An accounting firm has three departments. Each department consists of account clerks, executives and accountants. The numbers are as follows;

Department 1

Accountants - 1

Executives - 2

Account clerks -.8

Department 2

Accountants - 3

Executives - 5

Account clerks - 14

Department 3 

Accountants - 2

Executives - 3

Account clerk - 12

If five workers are selected at random, find:

i) the number of ways at least three workers from Department 3 were chosen.

ii) the probability that two executives were chosen.

Answer 1

i) The total number of workers in Department 1$:1+2+8=11.$

The total number of workers in Department 2$:3+5+14=22.$

The total number of workers in Department 1$:2+3+12=17.$

The number of workers in Department 1 and Department 2: $11+22=33$

The number of ways at least three workers from Department 3 were chosen is

$$\dbinom{17}{3}\dbinom{33}{5-3}+\dbinom{17}{4}\dbinom{33}{5-4}+\dbinom{17}{5}\dbinom{33}{5-5}$$

$$=680(528)+2380(33)+6188(1)=443768$$

(ii) The total number of workers: $11+22+17=50$

The total number of  executives: $2+5+3=10$

The probability that two executives were chosen is

$$\dfrac{\dbinom{10}{2}\dbinom{50-10}{5-2}}{\dbinom{50}{5}}=\dfrac{45(9880)}{2118760}$$

$$=\dfrac{11115}{52969}\approx0.2098$$