Vtyjkyuk

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.

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 $defPrmathsf PPr(A),Pr(B),Pr(C),Pr(D)$ 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 $$binom 41Pr(A)-binom 42Pr(A,B)+binom 43Pr(A,B,C)-binom 44Pr(A,B,C,D)$$

To do: Evaluate those probabilities.

Okay, one is $mathsf P(A,B)=dfrac1binom 82binom 62$

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

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 = $frac34$
Probability that $B$ gets wrong pair of gloves = $frac23$
Probability that $C$ gets wrong pair of gloves = $frac12$
And finally $D$ will left with no choice.

Now, let $E$ = No child gets right pair of gloves
then,
$$P(E) = frac34×frac23×frac12$$
$$P(E) = frac14$$
Probability that atleast one child gets right pair of gloves =$ 1 - P(E) = frac34$