Respuesta :
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[tex]=\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}[/tex]
[tex]=\frac{15\times14\times13\times12}{24} =1365[/tex]
The number of ways 4 persons can be selected from total 12+15 = 27 persons is:
27C4[tex]=\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}[/tex]
[tex]=\frac{27\times26\times25\times24}{24} =17550[/tex]
Therefore, the probability of randomly selecting four person committee consisting entirely of women from a pool of 12 men and 15 women is:
[tex]\frac{\binom{15}{4}}{\binom{27}{4}} =\frac{1365}{17550}=0.0778[/tex]
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.
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
[tex]\\ =\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!}[/tex]
[tex]=\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}[/tex]
[tex]=\frac{32760}{421200}[/tex]
[tex]=\frac{7}{90}=0.07777[/tex]
Therefore, probability of randomly selecting four person committee consisting entirely of women from a pool of 12 men and 15 women is 0.0777.
