Suppose that a race car driver will drive in 5 races and has a 30% probability of winning in each race. Assume that his performances in these races are independent. The driver signs a contract with an advertiser. The advertiser will pay the driver a bonus of $5,000 if he wins three races, $12,000 if he wins four races, and $20,000 if he wins all five. But the driver will pay the advertiser $5,000 if he can not win more than two races. What is the probability of collecting more than $5,000 of the bonus?

Question

BlackTom AI · Accepted Answer

To determine the probability of collecting more than $5,000 of the bonus, first interpret the payout structure. The driver gets $5,000 for exactly 3 wins, $12,000 for exactly 4 wins, and $20,000 for all 5 wins. The condition 'more than $5,000' excludes exactly 3 wins, since that payout is exactly $5,000. Therefore, we need the probability of winning either 4 or 5 races.

Option 1 (0.162): This would be far too large for the probability of 4 or 5 successes with p = 0.3 in 5 trials; it overestimates the likelihood of getting almost all wins.

Option 2 (0.028): This value matches the probability of exactly 4 wins, but it ignores the possibility of winning all 5 races, which also yields more than $5,000 in bonus.

Option 3 (0.03): This is the sum of the probabilities of exactly 4 wins and exactly 5 wins, rounded appropriately. Specifically:
- P(X = 4) = C(5,4) (0.3)^4 (0.7)^1 = 5 × 0.0081 × 0.7 = 0.02835
- P(X = 5) = (0.3)^5 = 0.00243
- Total P(X ≥ 4) = 0.02835 + 0.00243 ≈ 0.03078 ≈ 0.03 when rounded to two decimal places in probability terms.
Thus, this option captures both scenarios that yield more than $5,000.

Option 4 (0.132): This value is much larger than the true P(X ≥ 4) for p = 0.3 with n = 5; it may correspond to a much higher success count probability and does not align with binomial probabilities here.

Option 5 (0.057): This is closer to the probability of exactly 4 wins alone (0.02835) plus a portion of another outcome, but it does not precisely reflect the binomial sum for X ≥ 4 with p = 0.3. The correct total for 4 or 5 wins is about 0.0308, not 0.057.

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