A box contains $2$ red balls, $3$ green balls and $x$ blue balls, where \(x\in \mathbb{Z}

Question

A box contains $2$ red balls, $3$ green balls and $x$ blue balls, where $x\in \mathbb{Z}$, $x\ge 5$. A game is played where the contestant picks $5$ balls from the box without replacement. The total score, $S$, for the contestant is the sum of the number of green balls chosen and thrice the number of red balls chosen. The blue balls will not contribute any points, unless all $5$ balls are blue. If all the $5$ balls are blue, the score will be $25$ points. Show that $\mathrm{P}\left( S=6 \right)=\frac{20x\left( {{x}^{2}}-3x+14 \right)}{\left( x+5 \right)\left( x+4 \right)\left( x+3 \right)\left( x+2 \right)\left( x+1 \right)}$.[2] Given that $\mathrm{P}\left( S=6 \right)=\frac{5}{63}$, calculate $x$.[2] Complete the probability distribution table for $S$.[4] $s$12345678925 P${(S = s)}$${\frac{5}{42}}$${\frac{5}{63}}$${\frac{5}{21}}$${\frac{5}{63}}$${\frac{5}{84}}$${\frac{1}{252}}$ Evaluate $\mathrm{E}\left( S \right)$ and find the probability that $S$ is more than $\mathrm{E}\left( S \right)$.[2] Find the probability that there are no green balls drawn given that $S$ is more than $\mathrm{E}\left( S \right)$.[2]

Accepted Answer

(ii) $x = 5$, (iv) $E(S) = \frac{1159}{252}$, $P(S > E(S)) = \frac{127}{252}$, (v) $\frac{11}{127}$

2017 CJC P2 Q10

2017 CJC P2 Q10

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