In a tournament, a team plays 10 matches with probabilities of winning and losing each match as 1/3 and 2/3 respectively. Let x be the number of matches that the team wins, and y be the number of matches that team loses. If the probability P(|x-y| ≤ 2) is p, then 3⁹ p equals _________.

Solution

Maths sir · 4 steps

Since x + y = 10, the condition on |x - y| is really a condition on x alone.

1. One variable only. Every match is either won or lost, so y = 10 - x. Then |x - y| = |2x - 10| = 2|x - 5|.
2. Find allowed wins. 2|x - 5| ≤ 2 means |x - 5| ≤ 1, so x = 4, 5 or 6. Here x is binomial with n = 10, P(win) = 1/3 and P(loss) = 2/3.

   p = (C(10, 4)2⁶ + C(10, 5)2⁵ + C(10, 6)2⁴)/3¹⁰

3. Add the terms. 210 × 64 = 13440, 252 × 32 = 8064 and 210 × 16 = 3360. Their sum is 24864, so p = 24864/3¹⁰.
4. So. 3⁹ p = 24864/3 = 8288.

The common slip: 24864. Stopping at the numerator over 3¹⁰ and reporting it gives 24864. The question asks for 3⁹ p, so divide by 3 once more.

The idea to keep. When two counts add up to a fixed total, rewrite a condition on both as a condition on one. Then it is a plain binomial sum over a few values.

Answer from the NTA final answer key. Solution by PrepWiser.