2018 TJC P2 Q8

2018 TJC P2 Q8

9 marks

Box \(A\) contains five cards numbered \(1\), \(1\), \(2\), \(2\) and \(3\) and Box \(B\) contains three cards numbered \(4\), \(4\) and \(5\). Cards numbered \(1\) and \(5\) are red, whereas cards numbered \(2\), \(3\) and \(4\) are blue.

A card is drawn from each of the two boxes. If both cards are the same colour then the score will be the product of the numbers on the two cards. If both cards are different in colour then the score will be the sum of the numbers on the two cards.

Let \(X\) be the score obtained. The probability distribution of \(X\) is given below:

\(x\)\(5\)\(7\)\(8\)\(12\)
\(P(X = x)\)\(h\)\(k\)\(\frac{1}{3}\)\(k\)
  1. Show that \(h = \frac{2}{5}\).[2]
  2. Find \(k\), \( \mathrm{E}(X)\) and \( \mathrm{Var}(X)\).[4]
  3. Find the probability that the mean score of \(60\) independent observations of \(X\) lies between \(6.5\) and \(7.5\).[3]

In a game, a player pays \(\$20\) and draws a card each from Box \(A\) and Box \(B\). She receives, in dollars, an amount equal to three times the score obtained. Find the variance of her winnings per game.

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Answer:(i) \(h=\frac{2}{5}\) (ii) \(k=\frac{2}{15}\), \(E(X)=\frac{36}{5}\), \(\operatorname{Var}(X)=\frac{392}{75}\) (iii) \(0.836\) (iv) \(47.04\)

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