The Raffle and the Quality Check
A charity raffle has 200 tickets. Priya buys 5 tickets. There are 3 prizes, drawn one at a time without replacement, and a ticket cannot win twice. Separately, a factory produces components in batches of 20, and historical records show 5% are defective. An inspector tests 4 components chosen at random from a batch without replacement and accepts the batch only if all 4 pass. A colleague says: "Since only 5% are defective, you can just treat the 4 tests as independent and use 0.95 to the power 4."
1. Calculate the probability that Priya wins at least one prize, showing your method.
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Use the complement: P(at least one) = 1 − P(no prize).
Method 1 - direct product. Priya fails to win the first prize with probability 195/200. Given that, 195 non-Priya tickets remain of 199, so she fails the second with probability 195/199, and the third with 195/198.
P(no prize) = (195/200) × (195/199) × (195/198) = 0.975 × 0.97990 × 0.98485 ≈ 0.9408
So P(at least one prize) ≈ 1 − 0.9408 = 0.0592, about 5.9%.
Method 2 - combinations. Number of ways to choose 3 winning tickets from 200 is C(200,3). Ways to choose all 3 from the 195 non-Priya tickets is C(195,3).
P(no prize) = C(195,3)/C(200,3) = (195×194×193)/(200×199×198) ≈ 0.9408, giving the same answer.
Note that 3/200 = 1.5% would be badly wrong; the chance of at least one win is much larger than the chance of winning any specified single draw, because there are three chances.
2. Evaluate the colleague's independence shortcut for the inspector's test, and explain when it would be acceptable.
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The shortcut gives 0.95⁴ ≈ 0.8145, but it is not exactly right, because sampling without replacement from a batch of 20 makes the tests dependent: each pass removes a good component and slightly raises the defect proportion among those left.
Exact hypergeometric calculation. With 5% of 20 defective, the batch has 1 defective and 19 good.
P(all 4 pass) = (19/20) × (18/19) × (17/18) × (16/17) = 16/20 = 0.80
The intermediate terms cancel, leaving simply the proportion of good components remaining after four draws.
Comparison: 0.80 versus 0.8145 - a difference of about 1.5 percentage points. Small here, but it grows as the sample becomes a larger fraction of the batch.
When the shortcut is acceptable: when the sample is a small fraction of the population - the usual rule of thumb is under about 10% - so that removing items barely changes the composition. For a batch of 20 sampled 4 at a time, that is 20% of the batch, which is too large a fraction to ignore. For a batch of 2,000 sampled 4 at a time, the binomial approximation would be excellent.
3. The inspector wants to increase confidence in accepted batches. Compare doubling the sample to 8 with testing the same 4 but rejecting the batch if any single component fails.
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Doubling the sample to 8. With 1 defective in 20, P(all 8 pass) = 12/20 = 0.60. The probability of catching a defective batch rises from 0.20 to 0.40. The cost is doubled: eight tests instead of four, and more good batches wrongly rejected.
Rejecting on any single failure with 4 tests. This is already what the stated rule does - the batch is accepted only if all 4 pass. So this option is not a change; it is the current procedure, with a 0.20 chance of detecting the defect. To make it stricter you would have to change the acceptance number, not the sample size.
Better framing: the real choice is between spending more tests (larger n) and tightening the rule (accept only if all pass, which is already the strictest rule for a given n). A sensible compromise is a larger sample with a small allowed number of failures, which controls both the producer's risk of rejecting good batches and the consumer's risk of accepting bad ones. The trade-off should be stated in those terms, with the two error probabilities quantified.