Conditional Probability

What is Conditional Probability?

The probability for the occurrence of events that have some kind of condition is a conditional probability. Here one of the events is usually dependent on the other event. The event happening later is dependent on the occurrence of the first event So to find the probability of the latter event the probability of the first event. Hence the first event is a normal probability, but the second event is the conditional probability.

So event A is considered as a condition for event B to happen. Or in other words, the probability of event B is dependent on the probability of event A. We denote this conditional probability as \(P(B|A)\).

Also, note that it is not stated that both events occur simultaneously nor do both the events always have casual relationships.

How to calculate conditional probability

There is a formula we can use to find the probability of B given A has already occurred:

\[P(B|A) = \frac{P(A\cap B)}{P(A)}\]

Where:

P(B|A) is the probability of B given A

P(A∩B) is the probability of both A and B occurring and

P(A) is the probability of A occurring

Conditional probability tree diagram

A Tree Diagram can be a useful way to visualise and solve problems that contain conditional probabilities. What we need to do is draw the first two branches for event A and then the 4 branches for event B.

For example, let's imagine we had a bag containing 10 sweets that were either strawberry or lemon flavoured. We then picked one random sweet from the bag, ate it and then picked another one. If we knew that there were 6 strawberry sweets at the start, we could start drawing a tree diagram showing the probabilities of picking either lemon or strawberry sweets. The first time we picked, the probability of picking a strawberry sweet would be 6/10 or 0.6 and hence the probability of picking a lemon sweet has to be \(1-0.6=0.4\). From this, we can draw the first branches of our tree diagram.

First branches of tree diagram showing the probability of picking a strawberry or lemon sweet from a bag

Now, what happens for the second sweet we pick? Remember that the first sweet we picked was not put back into the bag so the total number of sweets in the bag is now 9 and the flavour picked on the second draw is dependent on the flavour picked in the first draw. If on the first pick, we took a strawberry sweet, there will only be 5 strawberry sweets left in the bag so the probability of picking a strawberry sweet now would be 5/9=0.556 and the probability of picking a lemon sweet would be \(1 - 0.556= 0.444\)

However, if in the first pick we took a lemon sweet, there will now be 6 strawberry and 3 lemon sweets left. So the probability of picking a strawberry sweet in this situation is \(\frac{6}{9} =0.667\) and the probability of picking a lemon sweet is \(1-0.667 = 0.333\). We can now draw the next four branches in our tree diagram:

Tree diagram showing the probability of picking 2 sweets from a bag containing strawberry and lemon sweets

The four branches we drew represent the conditional probabilities of different events. The first one from the top gives the probability of picking strawberry (S) given that strawberry was already picked the first time, so \(P(S_2 | S_1) = 0.556\). This same logic can be applied to all the following branches, giving \(P(L |S) = 0.444, \space P(S|L) = 0.667 \text{ and } P(L_2 | L_1) = 0.333\).

Conditional Probability Properties

The conditional probability properties mentioned here are all based on the formula above.

Here P(S) is the probability of the sample space and P(A) and P(B) are the probabilities of events A and B respectively.

• \(P(S|A) = P(A|A) = 1\)

• If P(X) is any event such that \(P(E) ≠ 0\), then \(P((A \cup B)|X) = P(A|X) + P(B|X) - P((A \cap B)|X)\)

• \(P(B'|A) = 1 -P(B|A)\). Here B’ is the complement set of B.

Conditional probability Venn diagram

Venn diagrams are another method we can use to solve conditional probability problems. To draw a Venn diagram, we need to know the probability of event A, the probability of event B and the probability of A and B. For example, a survey was conducted on 65 people asking about the flavours of ice cream they liked. 30 people said they like only chocolate (C), 20 people said they liked only vanilla (V), 10 people liked both and the rest liked neither.

From this, we know that the probability that someone only likes chocolate is \(\frac{30}{65}=0.462\). The probability that someone likes both vanilla and chocolate is \(P(C\cap V)=\frac{10}{65}=0.154\). The probability that someone likes neither is \(65-\frac{(30+20+10)}{65} = 0.0769\). We can draw the following Venn diagram:

Venn diagram showing the probability of people liking chocolate and vanilla ice cream.

Note that adding all the probabilities on a Venn diagram should always equal 1.

From this, we can use the formula \(P(B |A) = \frac{P(A \cap B)}{P(A)}\) to find the probability that someone likes chocolate ice cream given that they like vanilla:

\(P(C |V) = \frac{P(C \cap V)}{P(V)}\).

We already know that \(P(C\cap V)= 0.154\). P(V) will be the sum of the probabilities of someone liking just vanilla and someone likes both vanilla and chocolate. \(P(V)=0.308+0.154=0.462\). So,

\(P(C|V) = \frac{0.154}{0.462} = 0.333\).

Bayes theorem for conditional probability

Bayes theorem states that \(P(B|A) = \frac{P(A|B)}{P(A)}P(B)\)

This theorem can be proved by using the equations we used before. We know that \(P(B|A) = \frac{P(A \cap B)}{P(A)}\) and \(P(A|B) = \frac{P(A \cap B)}{P(B)}\).

If we rearrange both equations to find expressions for \(P(A \cap B)\), we get \(P(A \cap B) = P(B|A)P(A)\) and \(P(A \cap B) = P(A|B)P(B)\) . We can now equate the right-hand side of both of these expressions, so \(P(B|A)P(A) = P(A|B)P(B)\), therefore, \(P(B|A) = \frac{P(A|B)}{P(A)} P(B)\)