The events are not mutually exclusive because there is a 4 of clubs (both a 4 and of the club suit). So the probability we want would be
[tex]\mathbb P(\text{4 or club})=\mathbb P(\text{4})+\mathbb P(\text{club})-\mathbb P(\text{4 and club})[/tex]
There are four 4's in the deck, so the probability of drawing a 4 is
[tex]\mathbb P(\text{4})=\dfrac{\binom41\binom{48}0}{\binom{52}1}=\dfrac4{52}=\dfrac1{13}[/tex]
(I use the binomial coefficient here just to illustrate how we're counting the number of ways to draw a 4. There are four possible ways to draw one of them, i.e. 4 choose 1. On the other hand, we are not drawing any of the other 48 remaining cards, i.e. 48 choose 0. In the denominator, we're drawing 1 card from 52, i.e. 52 choose 1.)
There are 13 clubs in total, so the probability of drawing just 1 one of them is
[tex]\mathbb P(\text{club})=\dfrac{\binom{13}1\binom{39}0}{\binom{52}1}=\dfrac{13}{52}=\dfrac14[/tex]
There is only one 4 of clubs in the deck. The probability of drawing it is
[tex]\mathbb P(\text{4 and clubs})=\dfrac{\binom11\binom{51}0}{\binom{52}1}=\dfrac1{52}[/tex]
So the probability we care about is
[tex]\mathbb P(\text{4 or clubs})=\dfrac1{13}+\dfrac14-\dfrac1{52}=\dfrac4{13}[/tex]
