N2018 P2 Q8

N2018 P2 Q8

Junior College 2
11 marks
Free

A bag contains \((n+5)\) numbered balls. Two of the balls are numbered \(3\), three of the balls are numbered \(4\) and \(n\) of the balls are numbered \(5\). Two balls are taken, at random and without replacement, from the bag. The random variable \(S\) is the sum of the numbers on the two balls taken.

  1. Determine the probability distribution of \(S\).[4]
  2. For the case where \(n=1\), find \(P(S=10)\) and explain this result.[1]
  3. Show that \(E(S)=\frac{10n+36}{n+5}\) and \(\operatorname{Var}(S)=\frac{g(n)}{(n+5)^2(n+4)}\) where \(g(n)\) is a quadratic polynomial to be determined.[6]

Default solution

(i) Let \(D=\binom{n+5}{2}\). The counts for \(S=6,7,8,9,10\) are \(1,6,2n+3,3n,\binom n2\), respectively. Divide each count by \(D\).

(ii) \(n=1:\quad P(S=10)=0\), since only one ball is numbered \(5\).

(iii) \(E(S)=\frac{6+7(6)+8(2n+3)+9(3n)+10\binom n2}{D}=\frac{10n+36}{n+5}\).

\(E(S^2)=\frac{6^2+7^2(6)+8^2(2n+3)+9^2(3n)+10^2\binom n2}{D}=\frac{100n^2+642n+1044}{(n+5)(n+4)}\).

\(\operatorname{Var}(S)=E(S^2)-[E(S)]^2=\frac{22n^2+78n+36}{(n+5)^2(n+4)}\).

Answer:(i) For \(S=6,7,8,9,10\), the respective probabilities are \(1/D,\ 6/D,\ (2n+3)/D,\ 3n/D,\ \binom n2/D\), where \(D=\binom{n+5}{2}\); (ii) \(0\); (iii) \(g(n)=22n^2+78n+36\)

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