Compute the Probability that a hand of 3 cards will contain a king and an ace of the same suit

Question

This is a problem for my Homework and I see Inclusion-Exclusion method but I still cannot understand it. Can someone please help me. Thanks

EDIT:

I tried doing it this way, and I am not sure if this is the correct way to do it

(13C2 * 39C1)/52C3

Answer 1

First thought: by drawing $3$ cards it cannot happen that of two different suits a king and an ace are drawn. So it is enough to find out the probability that this happens for suit diamonds and to multiply the result by $4$ on base of mutually exclusiveness and symmetry.

So draw $3$ cards and let $D$ denote the event that king and ace of diamonds are selected.

Let $E_i$ denote the event that the $i$-th card drawn is the king or the ace of diamonds.

Then $D=(E_1\cap E_2)\cup(E_1\cup E_3)\cup(E_2\cup E_3)$ is a union of $3$ equiprobable and mutually exclusive events so that:

$$P(D)=3\times P(E_1\cap E_2)=3\times P(E_1)P(E_2\mid E_1)=3\times\frac2{52}\frac1{51}$$

This leads to: $$P(\text{king and ace of same suite are selected})=4\times3\times\frac2{52}\frac1{51}$$

On base of mutually exclusiveness.