HMMT 十一月 2023 · 冲刺赛 · 第 12 题

HMMT November 2023 — Guts Round — Problem 12

[8] A jar contains 97 marbles that are either red, green, or blue. Neil draws two marbles from the jar 5 without replacement and notes that the probability that they would be the same color is . After Neil 12 puts his marbles back, Jerry draws two marbles from the jar with replacement. Compute the probability that the marbles that Jerry draws are the same color. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2023, November 11, 2023 — GUTS ROUND Organization Team Team ID#

解析

[8] A jar contains 97 marbles that are either red, green, or blue. Neil draws two marbles from the jar 5 without replacement and notes that the probability that they would be the same color is . After 12 Neil puts his marbles back, Jerry draws two marbles from the jar with replacement. Compute the probability that the marbles that Jerry draws are the same color. Proposed by: Rishabh Das 41 Answer: 97 5 40 · 97 Solution 1: Note that = . Of all of the original ways we could’ve drawn marbles, we are 12 97 · 96 adding 97 ways, namely drawing the same marble twice, all of which work. Thus, the answer is 40 · 97 + 97 41 = . 97 · 96 + 97 97 Solution 2: Let there be a , b , and c marbles of each type. We know a + b + c = 97 and 2 2 2 2 2 2 a − a + b − b + c − c a + b + c − 97 5 40 = = = . 97 · 96 97 · 96 12 96 2 2 2 This means a + b + c = 41 · 97. Then the probability Bob’s marbles are the same color is 2 2 2 a + b + c 41 = . 2 97 97 Remark. A triple ( a, b, c ) that works is (12 , 32 , 53).