Probability: Bayes; medical screening — HCI 2025 H2 Math Prelim Paper 2
What this question tests
Question
A basketball free throw game involves a team of two players, Ben and John. The game consists of at most three throws and the moment a player makes a successful shot, the team wins, and the game will end.
The game uses the following rules.
- Only one player is selected for each game and the probability that Ben is selected in a game is \(0.7\).
- The probabilities that Ben and John make a successful shot in any single attempt are \(0.1\) and \(0.07\) respectively.
- The shots are independent of each other.
(a) Find the probability that the team wins the game.
(b) The team did not win the game. Find the probability that John was the one who was selected to shoot.
(c) The team attempts the game repeatedly until the first game is won. Find the least number of attempts required such that the probability of winning within \(n\) games is at least \(0.95\).
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(a) The team wins iff the selected player makes a successful shot within at most 3 attempts. \[\begin{aligned} \mathrm{P}(\text{wins}) &= \mathrm{P}(\text{wins}\mid\text{Ben})\mathrm{P}(\text{Ben})+\mathrm{P}(\text{wins}\mid\text{John})\mathrm{P}(\text{John}) \\ &= \bigl(0.1+0.9\times 0.1+0.9^{2}\times 0.1\bigr)(0.7) \\ &\phantom{=}+\bigl(0.07+0.93\times 0.07+0.93^{2}\times 0.07\bigr)(0.3) \\ &= 0.2483929. \end{aligned}\]
Alternative. \[ \mathrm{P}(\text{wins})=(1-0.9^{3})(0.7)+(1-0.93^{3})(0.3)=0.2483929. \]
(b) \[\begin{aligned} \mathrm{P}(\text{John selected}\mid\text{did not win}) &= \frac{\mathrm{P}(\text{John selected and did not win})}{\mathrm{P}(\text{did not win})} \\ &= \frac{0.3\times 0.93^{3}}{1-0.2483929} \\ &= 0.321\ (\text{to 3 s.f.}). \end{aligned}\]
(c) Let the game be played up to \(n\) times. \[ \mathrm{P}(\text{win within }n\text{ games}) = 1-(1-0.2483929)^{n}. \]
Alternative. \[\begin{aligned} \text{Required probability} &= \mathrm{P}(\text{win in 1st game})+\mathrm{P}(\text{win in 2nd game}) \\ &\quad+\mathrm{P}(\text{win in 3rd game})+\cdots+\mathrm{P}(\text{win in }n^{\text{th}}\text{ game}) \\ &= 0.2483929+(1-0.2483929)(0.2483929) \\ &\quad+(1-0.2483929)^{2}(0.2483929)+\cdots+(1-0.2483929)^{n-1}(0.2483929) \\ &= \frac{0.2483929\bigl(1-(1-0.2483929)^{n}\bigr)}{1-(1-0.2483929)} \\ &= 1-(1-0.2483929)^{n}. \end{aligned}\]
We require \[ 1-(1-0.2483929)^{n}\ge 0.95 \quad\Longleftrightarrow\quad (1-0.2483929)^{n}\le 0.05. \]
| \(n\) | \((1-0.2483929)^{n}\) |
|---|---|
| \(10\) | \(0.0575 \,>\, 0.05\) |
| \(11\) | \(0.0432 \,<\, 0.05\) |
Smallest \(n\) is \(11\).