Answer:
[tex](a)\dfrac{92}{117}[/tex]
[tex](b)\dfrac{8}{39}[/tex]
[tex](c)\dfrac{25}{117}[/tex]
Step-by-step explanation:
Number of Men, n(M)=24
Number of Women, n(W)=3
Total Sample, n(S)=24+3=27
Since you cannot appoint the same person twice, the probabilities are without replacement.
(a)Probability that both appointees are men.
[tex]P(MM)=\dfrac{24}{27}X \dfrac{23}{26}=\dfrac{552}{702}\\=\dfrac{92}{117}[/tex]
(b)Probability that one man and one woman are appointed.
To find the probability that one man and one woman are appointed, this could happen in two ways.
P(One man and one woman are appointed)[tex]=P(MW)+P(WM)[/tex]
[tex]=(\dfrac{24}{27}X \dfrac{3}{26})+(\dfrac{3}{27}X \dfrac{24}{26})\\=\dfrac{72}{702}+\dfrac{72}{702}\\=\dfrac{144}{702}\\=\dfrac{8}{39}[/tex]
(c)Probability that at least one woman is appointed.
The probability that at least one woman is appointed can occur in three ways.
P(at least one woman is appointed)[tex]=P(MW)+P(WM)+P(WW)[/tex]
[tex]P(WW)=\dfrac{3}{27}X \dfrac{2}{26}=\dfrac{6}{702}[/tex]
In Part B, [tex]P(MW)+P(WM)=\frac{8}{39}[/tex]
Therefore:
[tex]P(MW)+P(WM)+P(WW)=\dfrac{8}{39}+\dfrac{6}{702}\\$P(at least one woman is appointed)=\dfrac{25}{117}[/tex]