Conditional Probability (GATE-2012)

Suppose we assign a distribution function to a sample space and then learn that

an event E has occurred.How should we change the probabilities of the remaining

events? We shall call the new probability for an event F the conditional probability

of F given E and denote it by P (F |E).

lets try to understand the concept of conditional probability by taking few examples.

In the Life Table , one finds that in a population

of 100,000 females, 89.835% can expect to live to age 60, while 57.062% can expect

to live to age 80. Given that a woman is 60, what is the probability that she lives

to age 80?

This is an example of a conditional probability. In this case, the original sample

space can be thought of as a set of 100,000 females. The events E and F are the

subsets of the sample space consisting of all women who live at least 60 years, and

at least 80 years, respectively. We consider E to be the new sample space, and note

that F is a subset of E. Thus, the size of E is 89,835, and the size of F is 57,062.

So, the probability in question equals 57,062/89,835 = .6352. Thus, a woman who

is 60 has a 63.52% chance of living to age 80.

Consider our voting example : three candidates A,

B, and C are running for office. We decided that A and B have an equal chance of

winning and C is only 1/2 as likely to win as A. Let A be the event “A wins,” B

that “B wins,” and C that “C wins.” Hence, we assigned probabilities P (A) = 2/5,

P (B) = 2/5, and P (C) = 1/5.

Suppose that before the election is held, A drops out of the race.

it would be natural to assign new probabilities to the events B and C which

are proportional to the original probabilities. Thus, we would have P (B| A) = 2/3,

and P (C| A) = 1/3. It is important to note that any time we assign probabilities

to real-life events, the resulting distribution is only useful if we take into account

all relevant information. In this example, we may have knowledge that most voters

who favor A will vote for C if A is no longer in the race. This will clearly make the

probability that C wins greater than the value of 1/3 that was assigned above.

We have two urns, I and II. Urn I contains 2 black balls and 3 white

balls. Urn II contains 1 black ball and 1 white ball. An urn is drawn at random

and a ball is chosen at random from it. We can represent the sample space of this

experiment as the paths through a tree as shown in above Figure. The probabilities

assigned to the paths are also shown.

Let B be the event “a black ball is drawn,” and I the event “urn I is chosen.”

Then the branch weight 2/5, which is shown on one branch in the figure, can now

be interpreted as the conditional probability P (B|I).

Suppose we wish to calculate P (I|B). Using the formula, we obtain

P (I|B)=P (I ∩ B)/P(B)

=P (I ∩ B)/P (B ∩ I) + P (B ∩ II)

=(1/5)/(1/5 + 1/4)

=4/9