A committee of $10$ people is chosen at random from a group consisting of $18$ women and $12$ men. T

Question

A committee of $10$ people is chosen at random from a group consisting of $18$ women and $12$ men. The number of women on the committee is denoted by $R$. Find the probability that $R=4$. The most probable number of women on the committee is denoted by $r$. By using the fact that $\mathrm{P}\left( R=r \right)>\mathrm{P}\left( R=r+1 \right)$, show that $r$ satisfies the inequality $\left( r+1 \right)!\left( 17-r \right)!\left( 9-r \right)!\left( r+3 \right)!>r!\left( 18-r \right)!\left( 10-r \right)!\left( r+2 \right)!$ And use this inequality to find the value of $r$. Use the calculator to find $\mathrm{E}\left( R \right)$ and $\mathrm{Var}\left( R \right)$.

Accepted Answer

(i) $\frac{816}{8671}$; (ii) $r = 6$; (iii) $E(R) = 6$, $Var(R) = 1.66$

N2011 P2 Q11 [Modified]

N2011 P2 Q11 [Modified]

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