# How to Determine Independence of Events Using Probability

Let's get a deck of cards.

a) Are the events "the drawn card is a heart" and "the drawn card is an ace" independent?

b) Suppose we have removed the following cards from the deck: spades 2, 3, 4, and spade 5, and draw a card from the remaining 48. Since the removed cards are neither hearts nor aces, we could believe that the events in a) are still independent. Investigate if this is the case.

Soltion

a) P(A) = $$\frac{13}{52} = \frac{1}{4}$$ and $$P(B) = \frac{4}{52} = \frac{1}{13}$$. The event that the drawn card is a heart and an ace is $$A \cap B$$, and we have that $$P(A \cap B) = \frac{1}{52}$$.

b) $$P(A) = \frac{1}{52}$$ and $$P(B) = \frac{4}{48} = \frac{1}{12}$$. Finally, $$P(A \cap B) = \frac{1}{48}$$, but $$P(A \cap B) \neq P(A)P(B)$$.

I understand a) but on b) it is a little weird to me, how did they calculate $$P(A \cap B)$$ ? I mean it is not really clear to me why the second case is not independent either. I'd really appreaciate it if someone explain this.

• There are $48$ cards in total and exactly one of them is the $A\heartsuit$. But $P(A)$ is obviously not $\frac 1{52}$.
– lulu
Apr 15 at 10:32
• Look at en.wikipedia.org/wiki/Pearson%27s_chi-squared_test Mainly look at the example of a six-sided die. Apr 15 at 10:43
• There are 13 hearts out of 48 cards, so $P(A)=\frac{13}{48}\neq\frac{1}{52}$ as @lulu suggested above. Apr 15 at 10:43

Independent has a very specific meaning in probability, namely events $$A$$ and $$B$$ are independent if and only if $$Pr(A|B)=Pr(A)$$.

A consequence of this, which comes from the rule $$Pr(A|B)=\frac{Pr(A\cap B)}{Pr(B)}$$ is that $$A$$ and $$B$$ are independent if and only if $$Pr(A\cap B)=Pr(A)\times Pr(B)$$.

So, even though the events you describe in the natural language of part b above may appear independent, if they do not follow the strict rule for independence in the probability sense, they are not, by definition, independent.

If there is one Ace of Hearts in a deck of 48 cards, $$Pr(A\cap B)=\frac{n(A\cap B)}{n(U)}=\frac{1}{48}$$.
There are four aces left so $$Pr(A)=\frac{4}{48}$$ and there are 13 Hearts so $$Pr(B)=\frac{13}{48}$$.
However $$\frac{4}{48}\times\frac{13}{48}\neq\frac{1}{48}$$ so $$A$$ and $$B$$ are not independent.