Answer:
a.5005
b.[tex]\frac{1960}{5005}[/tex]
c.1/715
d.714/715
Step-by-step 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
[tex]nC_r[/tex]
Using the formula
We have n=15
r=6
Total number of ways=[tex]15C_6[/tex]
Total number of ways=[tex]\frac{15!}{6!9!}[/tex]
Using the formula
[tex]nC_r=\frac{n!}{r!(n-r)!}[/tex]
Total number of ways=[tex]\frac{15\times 14\times 13\times 12\times 11\times 10\times 9!}{6\times 5\times 4\times 3\times 2\times 1\times 9!}[/tex]
Total number of ways=5005
b. The probability of having exactly 3 men in the group
=[tex]\frac{8C_3\times 7C_3}{15C_6}[/tex]
Using the formula
Probability,[tex]P(E)=\frac{favorable\;cases}{Total\;number\;of\;cases}[/tex]
The probability of having exactly 3 men in the group=[tex]\frac{\frac{8!}{3!5!}\times \frac{7!}{3!4!}}{5005}[/tex]
=[tex]\frac{\frac{8\times 7\times 6\times 5!}{3\times 2\times 1\times 5!}\times \frac{ 7\times 6\times 5\times 4!}{3\times 2\times 1\times 4!}}{5005}[/tex]
=[tex]\frac{56\times 35}{5005}[/tex]
The probability of having exactly 3 men in the group
=[tex]\frac{1960}{5005}[/tex]
c. The probability of all the selected people in the group are women
=[tex]\frac{8C_0\times 7C_6}{5005}[/tex]
The probability of all the selected people in the group are women
[tex]=\frac{\frac{8!}{0!8!}\times \frac{7\times 6!}{6!1!}}{5005}[/tex]
The probability of all the selected people in the group are women
[tex]=\frac{7}{5005}=\frac{1}{715}[/tex]
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
[tex]=1-\frac{1}{715}[/tex]
[tex]=\frac{715-1}{715}[/tex]
[tex]=\frac{714}{715}[/tex]
The probability of having at least one man in the group [tex]=\frac{714}{715}[/tex]
