In a carnival lucky dip game, a game master places n consolation tickets, $m$ blank tickets, and o

Question

In a carnival lucky dip game, a game master places n consolation tickets, $m$ blank tickets, and one golden ticket into a box. A contestant taking part in the game would pay $\$1$ to draw two tickets from the box. They would then be awarded $\$1$ for each consolation ticket drawn, and $\$10$ for the golden ticket if it is drawn. Nothing is awarded for the blank tickets drawn. Let $\$W$ represent the total amount awarded to a contestant after one game.

Show that $\mathrm{P}\left( W=1 \right)=\frac{2nm}{\left( n+m+1 \right)\left( n+m \right)}$, and determine the probability distribution of $W$.[4]
By considering $\mathrm{E}\left( W \right)$, show that if the game master expects to make a profit from the lucky dip game, then $m-n>19$.[3]

The game master then decides to run the game with $10$ consolation tickets and $40$ blank tickets. He is also issued with carnival lucky draw tickets to give away, and gives each contestant $Y$ lucky draw tickets after the game, where $Y=\left| W-4 \right|$.

Find $\mathrm{E}\left( Y \right)$ and $\mathrm{Var}\left( Y \right)$.[2]

Accepted Answer

(a) ${\mathrm{P}(W=0)=\frac{m(m-1)}{(n+m+1)(n+m)}}$; ${\mathrm{P}(W=2)=\frac{n(n-1)}{(n+m+1)(n+m)}}$; ${\mathrm{P}(W=10)=\frac{2m}{(n+m+1)(n+m)}}$ (c) ${\mathrm{E}(Y)=3.70	ext{ (3 s.f)}}$; ${\mathrm{Var}(Y)=0.562	ext{ (3 s.f).}}$

2022 EJC P2 Q7

2022 EJC P2 Q7

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