This is about the probability of an event given some information. What follows assumes you know how to calculate basic probabilities (two posts ago) and the probability of an intersection of events (my last post).
Let’s start with an example. What is the probability of rolling a 6 on the roll of a die? From basic probability, we know that it is the number of ways to roll a 6 (only 1 way) divided by the number of total things that can happen (6). So the probability is 1/6. Now what if the die is rolled and a friend cheats by telling you that the number rolled is odd. Intuitively, you would say that the probability is 0 as 6 is an even number, so the additional information tells you that a 6 is not possible. The probability of a 6 and an odd number is 0 because the number of ways you can roll a 6 and an odd number is 0.
Now if the die is rolled and your friend says that the number rolled is even, what is the probability that a 6 was rolled? Intuitively, knowing that the number is even should increase the chances that a 6 was rolled. We can answer this using the basic probability formula: the number of ways to roll a 6 and an even number divided by the number of even numbers. Knowing that the number is even reduces the number of total things that can happen from 6 to 3. And the number of ways you can roll a 6 and an even number is 1. So the new probability, thanks to your friend, is 1/3.
As always, maths has notation for this. Let A and B be two events. Then the notation for “the probability of event A given (or on condition that) B occurred” is P(A|B). From the examples above, if A is the event of rolling a 6, and B is the event of rolling an odd number or it’s the event of rolling an even number, then the equation to calculate this is
\[P\left(A\middle|B\right)=\frac{n\left(A\cap B\right)}{n\left(B\right)}\]From my last two posts, remember that n(something) means the number of ways that something can occur, and the symbol ∩ means intersection or “and”.
This equation can be shown to be equivalent to
\[P\left(A\middle|B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}\]where the probabilities are used instead of the numbers. This can be rearranged to give what is called the multiplication rule of probability
\[P\left(A\cap B\right)=P\left(A\middle|B\right)P\left(B\right)\]So if P(A ∩ B) = 0.3 and P(B) = 0.7, then
\[P\left(A\middle|B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}=\frac{0.3}{0.7}=\frac{3}{7}\]Another example that shows conditional probability and the multiplication rule of probability in action is the following:
There is a bag with 10 marbles in it: 4 red and 6 blue ones. Two marbles are picked from the bag without replacing the first ball picked. “Without replacing” is important because the probability of picking the second ball’s color is affected by the first ball picked. If the first ball was replaced, the probability of the second ball’s color would not depend in the first ball’s color, that is, the two picks would be independent of each other.
So let’s look at some of the probabilities in this experiment. The probability that the first ball is red is P(R₁) =4/10 = 2/5. Now the probability of the second ball picked is dependent on that as there is 1 less red ball and 1 less ball in total. So the probability that the second ball is red is P(R₂|R₁) = 3/9 = 1/3 because there are only 3 red balls left out of the 9 balls left. Likewise, P(B₂|R₁) =6/9 = 2/3. For simple experiments like this, tree diagrams are often used to get a complete picture of all the possibilities:
The last column of combination probabilities uses the multiplication rule previously stated. Using tree diagrams like this, you can answer many questions about the experiment by adding these probabilities:
- What is the probability of picking just one red marble?
2. What is the probability of picking two marbles of the same color?
\[P\left({R_2\cap R}_1\right)+\ P\left({B_2\cap B}_1\right)=\frac{2}{15}+\frac{1}{3}=\frac{7}{15}\]3. What is the probability of picking at least one red marble?
\[P\left({R_2\cap R}_1\right)+P\left({B_2\cap R}_1\right)+P\left({R_2\cap B}_1\right)=\frac{2}{15}+\frac{4}{15}+\frac{4}{15}=\frac{10}{15}=\frac{2}{3}\]Note that since the last column includes all the possible ways this experiment can go, all of these probabilities add up to 1. So to answer question 3, a more efficient way to calculate the answer is to subtract the one possibility excluded from 1:
\[1-P\left({B_2\cap B}_1\right)=1-\frac{1}{3}=\frac{2}{3}\]Counting the number of ways an event can happen in simple experiments like this is easy to do in your head. But what about questions like “how many poker hands (5 cards) can be made from a standard deck of 52 cards?”. Not so easy. So next time, I will talk about how we can “count” large possibilities like this.