Consider of drawing one card from a deck of $52$. Prove that the events of a spade being drawn and an ace being drawn are independent events.

Let $A$ be the event that a spade is drawn and let $B$ be the event that an ace is drawn.

Then, $\text P(A) = 4/52 = 1/13$ and $\text P(B) = 4/52 = 1/13$.

How can I calculate $\text P(A\cap B)$? And how can I prove that these events are independent since the question specifically asked to prove that they are independent?

  • $\begingroup$ What card did you get in $A \cap B$? What is the probability that you drew that card? What is the definition of independence? $\endgroup$ Commented Jan 2, 2019 at 16:32
  • $\begingroup$ $P(A\cap B)$ is the probability that the card drawn is both an ace and a spade. $\endgroup$ Commented Jan 2, 2019 at 16:32
  • $\begingroup$ Independence is determined by $\text P(A\cap B)=\text P(A)\times\text P(B)$ $\endgroup$
    – TheSimpliFire
    Commented Jan 2, 2019 at 16:33
  • 2
    $\begingroup$ In a deck of 52 cards there should be more than four spades be present ... $\endgroup$ Commented Jan 2, 2019 at 16:33
  • $\begingroup$ Is there really a $1/13$ probability of drawing a spade? $\endgroup$ Commented Jan 2, 2019 at 16:33

1 Answer 1


You didn't compute the probability of event $A$ correctly. There are $13$ spades in a standard deck. So $P(A)=13/52=1/4$. Note that $A\cap B$ corresponds to drawing the ace of spades and hence $$ \frac{1}{52}=P(A\cap B)=\frac{1}{4}\times\frac{1}{13}=P(A)P(B) $$

  • $\begingroup$ Thank you so very much!!! $\endgroup$ Commented Jan 2, 2019 at 16:50

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