Answer:
Step-by-step explanation:
The probability of famy size [tex]M=m[/tex] is
[tex]P(M=m)=\frac{8-m}{28},m=1,2,...7[/tex]
Let [tex]N[/tex] members of family ride the roller coaster, then
[tex]P(N=n|M=m)=\frac{1}{m},n=1,2,...m=0[/tex]
otherwise
Required probability=
[tex]P(M=6|N=5)=\frac{P(N=5,M=6)}{P(N=5)}\\\\=\frac{P(N=5|M=6)P(M=6)}{P(N=5, M=5)+P(N=5, M=6)+P(N=5, M=7)}[/tex]
(Since [tex]P(N=n|M=m)=0[/tex] if n>m [tex]P(N=n,M=m)=0[/tex]
[tex]\frac{P(N=5|M=6)P(M=6)}{P(N=5|M=5)P(M=5)+P(N=5|M=6)P(M=5)+P(N=5|M=7)P(M=7)}\\\\=\frac{\frac{1}{6}\times \frac{2}{28}}{\frac{1}{5}\times \frac{3}{28}\times\frac{1}{6}\times\frac{2}{28}\frac{1}{7}\times\frac{1}{28}}=0.3097[/tex]