Question

Randomly selecting four person committee consisting entirely of women from a pool of 12 men and 15 women. The event of selecting a woman and the event of selecting a woman the next time are independent or dependent? Probability of randomly selecting four person committee consisting entirely of women from a pool of 12 men and 15 women is _____?

Answer 1

Solution: We have to find the probability of randomly selecting four person committee consisting entirely of women from a pool of 12 men and 15 women.

The number of ways 4 women can be selected from 15 women is:

15C4$=\binom{15}{4}=\frac{15!}{(15-4)!4!} =\frac{15!}{11! \times 4!} =\frac{15\times 14\times 13 \times 12\times11!}{11! \times 4\times 3 \times2 \times1}$

              $=\frac{15\times14\times13\times12}{24} =1365$

The number of ways 4 persons can be selected from total 12+15 = 27 persons is:

27C4$=\binom{27}{4}=\frac{27!}{(27-4)!4!}=\frac{27!}{23! \times 4!} =\frac{27\times26\times25\times24\times23!}{23! \times 4\times3\times2\times1}$

              $=\frac{27\times26\times25\times24}{24} =17550$

Therefore, the probability of randomly selecting four person committee consisting entirely of women from a pool of 12 men and 15 women is:

$\frac{\binom{15}{4}}{\binom{27}{4}} =\frac{1365}{17550}=0.0778$

The event of selecting a woman and the event of selecting a woman the next time are dependent because the probability of selecting a woman at first draw is not same as the probability of selecting a woman at second draw.

Answer 2

We are asked if the event of selecting a woman and the event of selecting a woman the next time are independent or dependent.

We can see that events are dependent because if one woman is already picked in the first draw....the probability of choosing a woman in the second draw will be different because number of woman is reduced.

The probability of randomly selecting four person committee consisting entirely of women from a pool of 12 men and 15 women will be

$\\ =\frac{_{4}^{15}\textrm{c}}{_{4}^{27}\textrm{c}}\\ \\ =\frac{\frac{15!}{(15-4)!4!}}{\frac{27!}{(27-4)!4!}}\\ \\ =\frac{\frac{15!}{11!\cdot 4!}}{\frac{27!}{23!\cdot 4!}}\\ \\ =\frac{15!\cdot 23!}{27!\cdot 11!}$

$=\frac{15\cdot 14\cdot 13\cdot 12\cdot 11!\cdot 23!}{27\cdot 26\cdot 25\cdot 24\cdot 23!\cdot 11!}\\ \\ =\frac{15\cdot 14\cdot 13\cdot 12}{27\cdot 26\cdot 25\cdot 24\cdot}$

$=\frac{32760}{421200}$

$=\frac{7}{90}=0.07777$

Therefore, probability of randomly selecting four person committee consisting entirely of women from a pool of 12 men and 15 women is 0.0777.