Question

Five salesman A, B, C, D and E of a company are considered for a three member trade
delegation to represent the company in an international trade conference. Construct the sample
space and find the probability that (i) A is selected (ii)A is not selected (iii) Either A or B (not both) is selected.

Answer 1

We can choose 3 salesmen out of 5 in $C^3_5=\frac{5!}{3!\cdot2!}=10$ ways. So we have 10 possible combinations of 3 elements out of 5, these combinations define our sample space.

(i) If one of them is A, then we can choose 2 more salesmen out of 4 remaining (A is already chosen). We can do this in $C^2_4=\frac{4!}{2!\cdot2!}=6$ ways. Thus, the probability, one of the salesmen is A equals 6/10=0.6.

(ii) If A is not selected, 3 salesmen out of 4 is chosen (A is now removed). It can be done in $C^3_4=\frac{4!}{3!\cdot1!}=4$ ways. The probability of this is 4/10=0.4.

(iii) If A is chosen, we should choose 2 more salesmen out of 4. But we can't choose B, so now we should choose 2 salesmen out of 3. There are $C^2_3=\frac{3!}{2!\cdot1!}=3$ ways to do this. The same situation will take place if we choose B without A. Thus, we have 3+3=6 ways to choose A or B (not both). So the probability of this situation is 6/10=0.6.