Question

My class consists of 8 men and 7 women. I want to pick a group of 6 people for research.

Write each answer using fraction as needed.
a. In how many different ways can I pick this group?
b. What is the probability of having exactly 3 men in the group?
c. What is the probability of all the selected people in group are women?
d. What is the probability of having at least one man in the group?

Answer 1

a.5005

b.
`(1960)/(5005)`

c.1/715

d.714/715

Explanation:

We are given that

Total men=8

Total women=7

Total people, n=8+7=15

r=6

a.

Combination formula:

Selection of r out of n people by total number of ways

`nC_r`

Using the formula

We have n=15

r=6

Total number of ways=
`15C_6`

Total number of ways=
`(15!)/(6!9!)`

Using the formula

`nC_r=(n!)/(r!(n-r)!)`

Total number of ways=
`(15* 14* 13* 12* 11* 10* 9!)/(6* 5* 4* 3* 2* 1* 9!)`

Total number of ways=5005

b. The probability of having exactly 3 men in the group

=
`(8C_3* 7C_3)/(15C_6)`

Using the formula

Probability,
`P(E)=(favorable\;cases)/(Total\;number\;of\;cases)`

The probability of having exactly 3 men in the group=
`((8!)/(3!5!)* (7!)/(3!4!))/(5005)`

=
`((8* 7* 6* 5!)/(3* 2* 1* 5!)* ( 7* 6* 5* 4!)/(3* 2* 1* 4!))/(5005)`

=
`(56* 35)/(5005)`

The probability of having exactly 3 men in the group

=
`(1960)/(5005)`

c. The probability of all the selected people in the group are women

=
`(8C_0* 7C_6)/(5005)`

The probability of all the selected people in the group are women

`=((8!)/(0!8!)* (7* 6!)/(6!1!))/(5005)`

The probability of all the selected people in the group are women

`=(7)/(5005)=(1)/(715)`

d. The probability of having at least one man in the group

=1- probability of all the selected people in group are women

The probability of having at least one man in the group

`=1-(1)/(715)`

`=(715-1)/(715)`

`=(714)/(715)`

The probability of having at least one man in the group
`=(714)/(715)`