Conditional Probabilities

Conditional probabilities occur when events occur in succession or when additional information is provided about event that has occured. When the occurance of event A affects the occurance of event B we say that the events are dependent and we speak in terms of the conditional probability of B given A.

P(B|A) is read the conditional probability of B given that A has occured.

Suppose one card is drawn from a standard deck of 52 cards and we are provided with additional information that the card drawn is a spade. What is the probability that the card is a queen? Using the symbolism of probability we write

P(queen|spade) = ?

The idea of a reduced sample space is used calculate a conditional probability. Before the card was drawn from the deck of 52 cards, the sample space consisted of all 52 cards. Once we are told that the card drawn is a spade, the sample space is reduced to 13 cards(e.g. the spades). Now the question becomes, of the 13 spades, how many are queens? The answer is of course, one, namely the queen of spades. So

P(queen|spade) = 1/13 = .0769.

The formula for computing conditional probabilities is:



Lets try another problem. Suppose a card is drawn from a standard deck of 52 cards and we are told that the card is a Jack. What is the probability that it is a Heart?

The solution: Since there are only 4 Jacks in a standard deck of cards, the reduced sample space is 4. Now of the 4 Jacks only one is a Heart. Therefore:

P(Heart|Jack) = 1/4 = 0.25.

Ok ,since we are on a roll, lets try a different problem.
A jar contains 4 red balls; 5 yellow balls; and 3 green balls. If two balls are drawn in succession without replacement, what is the probability that the second ball drawn is a red ball given that the first ball drawn was a yellow ball?

The solution: Since the first ball drawn and not replaced was a yellow ball, the reduced sample space is now 11 balls instead of the original 12 balls. Of the eleven balls remaining, 4 of them are red. Therefore

P(R|Y) = 4/11 = .3636.

If the balls in the previous problem were drawn with replacement, then the probability of getting a red ball on the second draw would not depend on the fact that a yellow ball was drawn on the first draw. So
If events A and B are independent then the probability of event B given A is the simple probability of event B

P(B|A) = P(B)