What Is The Probability Of Drawing A Heart Given That The Card Is Red Given That It Is A Heart

Conditional Probability

The conditional probability of an consequence B is the probability that the consequence will occur given the knowledge that an consequence A has already occurred. This probability is written P(B|A), notation for the probability of B given A. In the case where events A and B are independent (where event A has no effect on the probability of result B), the conditional probability of event B given result A is simply the probability of effect B, that is P(B).

If events A and B are not contained, and so the probability of the intersection of A and B (the probability that both events occur) is defined by
P(A and B) = P(A)P(B|A).

From this definition, the provisional probability P(B|A) is easily obtained by dividing by P(A):

Notation: This expression is just valid when P(A) is greater than 0.

Examples

In a card game, suppose a player needs to draw ii cards of the same suit in club to win. Of the 52 cards, there are 13 cards in each suit. Suppose showtime the player draws a center. Now the player wishes to describe a second heart. Since one heart has already been chosen, there are now 12 hearts remaining in a deck of 51 cards. So the conditional probability P(Draw second eye|Start card a centre) = 12/51.

Suppose an private applying to a higher determines that he has an lxxx% take chances of being accepted, and he knows that dormitory housing will just be provided for threescore% of all of the accustomed students. The chance of the student being accustomed and receiving dormitory housing is divers past
P(Accustomed and Dormitory Housing) = P(Dormitory Housing|Accepted)P(Accepted) = (0.60)*(0.fourscore) = 0.48.

To calculate the probability of the intersection of more 2 events, the conditional probabilities of all of the preceding events must be considered. In the case of three events, A, B, and C, the probability of the intersection P(A and B and C) = P(A)P(B|A)P(C|A and B).

Consider the college applicant who has determined that he has 0.lxxx probability of acceptance and that only threescore% of the accepted students volition receive dormitory housing. Of the accepted students who receive dormitory housing, lxxx% will accept at least one roommate. The probability of being accepted and receiving dormitory housing and having no roommates is calculated by:
P(Accepted and Dormitory Housing and No Roommates) = P(Accepted)P(Dormitory Housing|Accepted)P(No Roomates|Dormitory Housing and Accepted) = (0.80)*(0.60)*(0.20) = 0.096. The student has virtually a 10% risk of receiving a single room at the college.

Some other important method for calculating provisional probabilities is given by Bayes'southward formula . The formula is based on the expression P(B) = P(B|A)P(A) + P(B|A^c)P(A^c), which simply states that the probability of outcome B is the sum of the conditional probabilities of result B given that event A has or has not occurred. For independent events A and B, this is equal to P(B)P(A) + P(B)P(A^c) = P(B)(P(A) + P(A^c)) = P(B)(1) = P(B), since the probability of an event and its complement must always sum to 1. Bayes's formula is defined equally follows:

Example

Suppose a voter poll is taken in 3 states. In state A, fifty% of voters support the liberal candidate, in land B, 60% of the voters support the liberal candidate, and in state C, 35% of the voters support the liberal candidate. Of the total population of the three states, xl% live in state A, 25% live in state B, and 35% alive in land C. Given that a voter supports the liberal candidate, what is the probability that she lives in land B?

By Bayes's formula,

        P(Voter lives in land B|Voter supports liberal candidate) = P(Voter supports liberal candidate|Voter lives in state B)P(Voter lives in state B)/ 	(P(Voter supports lib. cand.|Voter lives in state A)P(Voter lives in state A) +  	 P(Voter supports lib. cand.|Voter lives in state B)P(Voter lives in state B) + 	 P(Voter supports lib. cand.|Voter lives in state C)P(Voter lives in state C))        = (0.60)*(0.25)/((0.50)*(0.twoscore) + (0.60)*(0.25) + (0.35)*(0.35))  = (0.15)/(0.20 + 0.15 + 0.1225) = 0.15/0.4725 = 0.3175.

The probability that the voter lives in state B is approximately 0.32.

For some more definitions and examples, run into the probability alphabetize in Valerie J. Easton and John H. McColl'southward Statistics Glossary v1.1.