probability at least one child finds their pair of gloves

Question

Their are 4 pairs of gloves, $8$ gloves in total. $4$ children whose gloves are in the pile randomly select $2$ gloves.

What would the probability be that at least one child selects his pair of gloves?

I'm not sure how to approach this problem.

Answer 1

Hint: Look for the case when none of the children find pair calculate the probability and subtract from 1

Read this : https://en.m.wikipedia.org/wiki/Derangement

Answer 2

Welcome Emily. Yes, PIE is the way to go.

Let $A,B,C,D$ be the events that Ada, Ben, Carl, and Danae find their own gloves. These events are, of course, dependent.

Then $\def\Pr{\mathsf P}\Pr(A),\Pr(B),\Pr(C),\Pr(D)$ are the probabilities that one particular child finds their glove (and they are all equal), $\Pr(A,B),\Pr(A,C),\Pr(A,D),\Pr(B,C),\Pr(B,D),\Pr(C,D)$ the probabities that two particular children do so (and are likewise equal to each other), and so on.

Thus by the Principle of Inclusion and Exclusion you seek $$\binom 41\Pr(A)-\binom 42\Pr(A,B)+\binom 43\Pr(A,B,C)-\binom 44\Pr(A,B,C,D)$$

To do: Evaluate those probabilities.

Okay, one is $\mathsf P(A,B)=\dfrac{1}{\binom 82\binom 62}$

Answer 3

Simple method is to first compute probability that nobody gets the right pair and then substarct it from 1. Before procedding i will assume that gloves are kept in pair. Suppose that threre are four child $A, B, C, D$. Then, Probability that $A$ gets wrong pair of gloves = $\frac{3}{4}$ Probability that $B$ gets wrong pair of gloves = $\frac{2}{3}$ Probability that $C$ gets wrong pair of gloves = $\frac{1}{2}$ And finally $D$ will left with no choice.

Now, let $E$ = No child gets right pair of gloves then, $$P(E) = \frac{3}{4}×\frac{2}{3}×\frac{1}{2}$$ $$P(E) = \frac{1}{4}$$ Probability that atleast one child gets right pair of gloves =$ 1 - P(E) = \frac{3}{4}$