Probability: Two-stage; conditional — TJC 2025 H2 Math Prelim Paper 2
What this question tests
Question
In a simple archery game, a player shoots an arrow at a target. The probability of hitting the target depends on the player's skill and the wind conditions for the day. The wind can either be good or bad, with the probability of a good wind day being 0.7.
According to past statistics:
- Dave has a 70% chance of hitting the target on a good wind day and a 60% chance on a bad wind day.
- Rafael has an 80% chance of hitting the target on a good wind day and a 50% chance on a bad wind day.
Dave and Rafael decide to compete in a game where each player shoots 2 arrows at the target. The player who hits the target more times wins the game.
- Find the probability that Dave wins the game on a good wind day.
It is now known that the probability Dave wins the game on a bad wind day is 0.39.
- Given that Dave wins the game, find the probability that the day was a bad wind day.
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(a) On a good wind day: Dave hits with \(p = 0.7\), Rafael hits with \(q = 0.8\) per shot. Each shoots 2 arrows. Dave wins if he hits more times than Rafael.
Required probability = P(Dave hits 2, Rafael hits 0 or 1) + P(Dave hits 1, Rafael hits 0): \[\begin{aligned} &= \left(\frac{7}{10}\right)^{\!2}\!\left[\binom{2}{0}\!\left(\frac{2}{10}\right)^{\!2}+\binom{2}{1}\!\left(\frac{8}{10}\right)\!\left(\frac{2}{10}\right)\right]+\binom{2}{1}\!\left(\frac{7}{10}\right)\!\left(\frac{3}{10}\right)\!\left(\frac{2}{10}\right)^{\!2}\\ &= \frac{49}{100}\left[\frac{4}{100}+\frac{32}{100}\right] + \frac{42}{1000}\cdot\frac{4}{100}\\ &= \frac{49}{100}\cdot\frac{36}{100}+\frac{168}{100000}\\ &= 0.1764 + 0.0168 = 0.1932 \end{aligned}\]
(b) By Bayes' theorem: \[\begin{aligned} \mathrm{P}(\text{bad} \mid \text{Dave wins}) &= \frac{\mathrm{P}(\text{bad})\cdot\mathrm{P}(\text{Dave wins}\mid\text{bad})}{\mathrm{P}(\text{good})\cdot\mathrm{P}(\text{wins}\mid\text{good})+\mathrm{P}(\text{bad})\cdot\mathrm{P}(\text{wins}\mid\text{bad})}\\ &= \frac{(0.3)(0.39)}{(0.7)(0.1932)+(0.3)(0.39)}\\ &= \frac{0.117}{0.13524+0.117} = \frac{0.117}{0.25224} \approx 0.464 \end{aligned}\]