Question

Exercise 1.A committee of five people is chosen randomly from four men and six women. Find the probability that:a) Exactly four women are on the committee:b) At least four women are on the committee:c) At most four women are on the committee:

Answer 1

Answer with explanation:

Given : Number of men = 4

Number of women = 6

Total people = $6+4=10$

Total number of ways to make a committee of five people from 10 persons :-

$^{10}C_{4}=\dfrac{10!}{4!(10-4)!}=210$

a) Number of ways to make committee that has exactly four women :

$^6C_4\times ^4C_1=\dfrac{6!}{4!(6-4)!}\times4=60$

The  probability that committee has exactly four women :

$\dfrac{60}{210}=\dfrac{2}{7}$

b) Number of ways to make committee that has at-least four women :

$^6C_4\times ^4C_1+^6C_5=\dfrac{6!}{4!(6-4)!}\times4+6=66$

The probability that committee at-least four women :

$\dfrac{66}{210}=\dfrac{22}{70}$

c) Number of ways that committee has more than 4 women :-

$^6C_5\times^4C_0=6$

The probability that committee has more than 4 women :-

$\dfrac{6}{210}$

Now, the  probability that committee has at most four women :-

$1-\dfrac{6}{210}=\dfrac{204}{210}=\dfrac{102}{105}$