A fair coin is tossed, and a card is drawn from a standard deck (52 cards). What is the probability that you tossed a head or drew an ace?

Let 'A' be the event of getting head from the coin toss and 'B' be the event of getting ace from the deck.

My approach is to use the sum rule, which is

P(A or B) = P(A) + P(B) - P(A ∩ B)

My question is

Since tossing a coin and drawing a card are two independent events, can we take take P(A ∩ B) as zero?

  • 5
    $\begingroup$ Check the definition of "independent". $\endgroup$
    – lulu
    Commented Dec 4, 2022 at 11:12
  • 6
    $\begingroup$ $P(A\cap B)$ takes value $0$ if $A$ and $B$ are mutually exclusive. That is not the same thing as being independent. $\endgroup$
    – drhab
    Commented Dec 4, 2022 at 12:07
  • 1
    $\begingroup$ You can view this as an experiment having $2*52=104$ outcomes. Then it becomes straightforward. $\endgroup$
    – John Douma
    Commented Dec 4, 2022 at 19:48

2 Answers 2


The complement event to tossing a head $\color{red}{\text{or}}$ drawing an Ace is :

Tossing a tail and not drawing an Ace.

Probability of complement event occuring is

$$\frac{1}{2} \times \frac{12}{13} = \frac{6}{13}.$$

Therefore, probability of original event is $1$ minus the probability of the complement event, which is

$$1 - \frac{6}{13} = \frac{7}{13}.$$

In general, the assumption that $A$ and $B$ are two independent events does not imply that $p(A \cap B) = 0.$ Simple counter example: Toss a coin twice. Each coin toss is an independent event. If you let $A$ denote the event that first coin toss is Heads and $B$ denote the event that second coin toss is Heads, you do not have that $p(A\cap B) = 0.$ That is, the probability of getting two consecutive Heads on two coin tosses is not zero.


Another way to look at this is that there are $2*52=104$ outcomes. Since there are four aces and for each ace we can get either a head or a tail, there are $8$ ways to get an ace. We can also get a head and for that head there are $48$ non-ace cards which gives us a total of $8+48=56$ ways to get either an ace or a head.

Therefore, the probability is $\frac{56}{104}=\frac{7}{13}$


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