Question

Twelve people (four from each of three families) participate in a raffle drawing at a fair. Three of the twelve are randomly chosen (no person is allowed to be chosen

more than once) as winners.

(a) Find the probability that exactly one of the winners comes from each of the families.

(b) Would it be unusual for all of the winners to come from the same family?

(c) Would your answer in (b) differ if people were allowed to win more than once?

Answer 1

a) Chosing three out of twelve randomly = $^{12}\textrm{C}_{3}$

For one winner from one family = $^{4}\textrm{C}_{1}$

Therefore, probability that exactly one of the winners comes from each of the families = $\frac{(^{4}\textrm{C}_{1})^3}{^{12}\textrm{C}_{3}}$

=16/55

b) For choosing one family out of three = ${^{3}\textrm{C}_{1}$

For all winners from one family = $^{4}\textrm{C}_{3}$

Probability that all the winners are from one family is = $\frac{(^{4}\textrm{C}_{3})*(^{3}\textrm{C}_{1})}{^{12}\textrm{C}_{3}}$ = 3/55

Hence, 3/55 is very less so it is unusual for all the wwinners to come from same family.

c) If winning more than once was allowed then, for all the winners from one family = ${^{4}\textrm{C}_{3}}+{^{4}\textrm{C}_{2}}+{^{4}\textrm{C}_{1}}$

Probability that all the winners are from one family is = $\frac{(^{4}\textrm{C}_{3}+^{4}\textrm{C}_{2}+^{4}\textrm{C}_{1})*(^{3}\textrm{C}_{1})}{^{12}\textrm{C}_{3}}$ = 3/22

So the probability now increased than the previous case.

Hope this helps!