2014 UNCO Math Contest II Problems/Problem 7

Problem

The Caterpillar owns five different matched pairs of socks. He keeps the ten socks jumbled in random order inside a silk sack. Dressing in the dark, he selects socks, choosing randomly without replacement. If the two socks he puts on his first pair of feet are a mismatched pair and the two socks he puts on his second pair of feet are a mismatched pair, then what is the probability that the pair he selects for his third set of feet is a mismatched pair?

Solutions

Solution 1

Let the colors be $A,B,C,D,E$ . Note that for the first four to be mismatched, there are $3$ possible cases:

(A,B), (C,D) | (A,B) (A,C) | (A,B) (A,B)

and the permutations of these letters. We will work case by case. Let $N_i$ be the number of situations of case $i$ , and let $P(i)$ be the probability that we get a mismatched pair in case $i$ .

Case 1: (A,B) , (C,D). No pair is completely utilized. We have: $\boxed{ N_1 = \binom{5}{4} \cdot 2^4 \cdot 4! = 1920 }$ To explain, $\binom{5}{4}$ is the number of ways to select which $4$ pairs we will have. The $4!$ counts for the ways I can arrange these socks in pairs. The tricky part is the $2^4$ .

Suppose I have 2 red socks and 2 blue socks in a bag, and I randomly draw. What is the probability that I get a mismatched pair? One might argue that, well, I can either draw a RR, a GG or a RG. So the probability is $\frac13$ right? No, it is not. Let us draw anything for the first, it is irrelevant. Now, what is the probability that I do not draw this same color? Well there is one copy of this color, and 2 copies of the other color, so the probability I draw the mismatched pair is $\frac23$ . The issue is that when we select randomly, although the colors are the same, they are treated as distinct objects. This is because each object in the bag is equally likely to be selected. That is why, when I select from the bag, I have two of the same color, say $A,A$ . I have to factor in $which$ one of these two I do select, and to factor that in, the $2^4$ comes in, because I have 4 such considerations. That is why, whenever we count the total number of cases, we will be treating the same colors as two different entities. This is only done because we are trying to compute the probability. If we were asked the total number of actual cases in which mismatch happens, we would not consider the same colored socks to be different entities. The $Random$ drawing part caused this.

Now, we compute the $probability$ of getting the $3^{rd}$ mismatched pair. There is only one complete pair that we could possibly draw. $P(pair) = \frac{\binom{2}{2}}{\binom{6}{2}} \implies P(mismatched) = 1 - P(pair) = \boxed{P(1) = \frac{14}{15}}$ This should be clear. There is only one pair, and since order $(AB, BA)$ doesn't matter, we have to chose 2 from 6.

Case 2: (A,B) (A,C). One Pair is completely utilized. The number of ways for this is: $\boxed{ N_2 = \binom{5}{1} \cdot \binom{4}{2} \cdot 2^2 \cdot 16 = 1920}$ This is straightforward too. First we pick which color has the pair complete, then we pick the other two colors, and then we permute such that the pair does not end up together.

There are 2 complete pairs in the remaining choices. Thus, probability of mismatch = $1 - \frac{2}{15} = \boxed{P(2) = \frac{13}{15}}$ .

Case 3: (A,B) (A,B). Two complete pairs are utilized. $\boxed{N_3 = \binom{5}{2} \cdot 16 = 160}$ We pick which two pairs we want, and then we avoid the cases where the same colors get paired up.

There are 3 complete pairs left, so the probability of a mismatch is $1 - \frac{3}{15} = \boxed{P(3) = \frac{12}{15}}$ .

Total number of cases where the first two are mismatched $N = N_1 + N_2 + N_3 = 1920 + 1920 + 160 = 4000$ . This means, that the probability of case 1 is $\frac{N_1}{N}\cdot P(1) = \frac{1920}{4000} \cdot \frac{14}{15}$ To understand this, note that the first fraction is the probability that we get to the first case, and the second fraction is the probability that we get to the mismatched pair, and we multiply these together to get the total probability of the first case. From here, it is clear that net probability $P = \frac{N_1}{N}\cdot P(1) + \frac{N_2}{N}\cdot P(2) + \frac{N_3}{N}\cdot P(3)= \frac{N_1P(1) + N_2 P(2) + N_3 P(3)}{N} = \boxed{\frac{112}{125}}$

Note

This problem is very very similar to https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_5

2014 UNCO Math Contest II (Problems • Answer Key • Resources)
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2014 UNCO Math Contest II Problems/Problem 7

Contents

Problem

Solutions

Solution 1

Note

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