# Statistics – Probability of Combined Events

I ended my last post showing the probability of picking a type of card from a standard deck of 52 cards. For example, if the event of interest, A, is picking a Jack, then the probability of picking a Jack from a shuffled deck of cards is

$P\left(A\right)=\frac{4}{52}=\frac{1}{13}$

because there are 4 ways to pick a Jack out of 52 cards. Now let’s consider probabilities of events like “picking a Jack or a Heart” or “a face card and a Heart”.

If we let events A be picking a Jack, B be picking a Heart, and C be picking a face card (Jack, Queen, or King), then the maths notation for these statements are

$P\left(A\cup B\right)=\mathrm{probability\ of\ picking\ a\ Jack\ or\ a\ Heart}$ $P\left(B\cap C\right)=\mathrm{probability\ of\ picking\ a\ face\ card\ and\ a\ Heart}$

The symbol “∪” stands for the union of two events, but in English, you can use the word “or”: AB = “A union B” or “A or B“. The symbol “∩” stands for the intersection of two events, but in English, you can use the word “and”: BC = “B intersection C” or “B and C“. These concepts are easily seen in a Venn diagram:

Circle A is the set of all Jacks and circle B is the set of all Hearts. Now the probability of picking a card from set A is 4/52. The probability of picking a card from set B is 13/52. You may be tempted so say that the probability of A or B is the sum of the two individual probabilities. But both of these probabilities include the Jack of Hearts so it is used twice. We have to subtract out this intersection of the two probabilities, so in maths notation:

$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$

This equation can be rearranged to show that the probability of the intersection of the two events is equal to the sum of the individual probabilities minus the probability of the union:

$P\left(A\cap B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cup B\right)$

These two equations are different forms of what is called the addition rule of probability.

So P(AB) = 4/52 +13/52 – 1/52 = 16/52, because P(AB) is the probability of a Jack and a Heart. Only one card satisfies this, the Jack of Hearts, so the probability of that is 1/52.

Now let’s define event D as picking a Diamond and consider the probability of picking a Heart and a Diamond, P(BD). This is clearly 0 as a card cannot be both suits. The associated Venn diagram looks like:

Events like this are called mutually exclusive, that is, you can pick one or the other, the picked card cannot be both. For mutually exclusive events:

$P\left(B\cup D\right)=P\left(B\right)+P\left(D\right)\ \mathrm{and}\ P\left(B\cap D\right) =0$

In my next post, I will discuss what is called conditional probabilities and explore the probability of picking a Jack given that the card is a Heart.