<p>There are 10 prizes, five A's, three B's, and two C's, placed in identical sealed envelopes for the top 10 contestants in a mathematics contest. The prizes are awarded by allowing winners to select an envelope at random from those remaining. When the 8th contestant goes to select the prize, the probability that the remaining three prizes are one A, one B and one C is</p>
Step-by-Step Solution
Key Concept: The probability that specific prizes remain for the 8th contestant depends only on what the first 7 contestants drew, not on the order of selection. We need to find the probability that exactly 4 A's, 2 B's, and 1 C have been drawn by the first 7 contestants.
<p><strong>Step 1:</strong> For the remaining 3 prizes (when 8th contestant arrives) to be one A, one B, and one C, the first 7 contestants must have drawn: 4 A's, 2 B's, and 1 C.</p><p><strong>Step 2:</strong> By symmetry principle, the probability that any specific selection of 7 prizes from 10 is drawn equals the probability of any other specific selection of 7 prizes. The number of ways to choose 4 A's from 5, 2 B's from 3, and 1 C from 2 is: C(5,4) × C(3,2) × C(2,1) = 5 × 3 × 2 = 30</p><p><strong>Step 3:</strong> The total number of ways to choose 7 prizes from 10 is: C(10,7) = 120</p><p><strong>Step 4:</strong> Therefore, the probability = 30/120 = 1/4</p><p>∴ Answer: <strong>1/4</strong></p>
Correct Answer: 4