## How to prepare for JEE mains Conditional Probability How to prepare for JEE mains Conditional Probability? We will explore in this article.

## Conditional Probability

As probability is linked to our ignorance over the results of experience, the fact that an event happens, may change the probability of the others. The process of making the clinical history, explore and make complementary tests illustrates this principle.

The probability that an event A happens if an event B has happened is called conditional probability and it is defined as:

p ( A | B ) = p ( A B ) / p (B)

if p ( B ) =/= 0

This definition is consistent, which means it complies with the probability axioms. When an event happens, the sample space changes, that is why probability changes. Sometimes it is easier to calculate the conditional probability having in mind this change in the sample space.

### Example 1.

Let us consider an urn that has four red pebbles and five white pebbles. Of the four red pebbles, two are solid and two are striped. For the five white pebbles, four are solid and only one is striped.

Let us suppose that we take one pebble out of the urn and, without we looking at it, someone tells us that the pebble is red. Knowing this, we ask ourselves, what is the probability that the pebble is striped?

We will consider A and B as groups of events, being the A events: “the pebble is striped” and the B events: “the pebble is red”. Obviously, without any previous information, P(A) = 3/9 = 1/3 y P(B) = 4/9. Nevertheless, since we know that the pebble is red, the probability that it is striped is 2/4 = ½, since two of the four red pebbles are solid and the other two are striped.

Let us observe that, when B happened, the sample space got smaller. In general, given an experiment and it’s associated sample space, we will like to determine how knowing that an event B happened affects the probability of A.

### Example 2.

In Example 1, P(B) = 4/9 and

p ( A | B ) = p ( A B ) / p (B) = [2/9] / [4/9] = 2/4 = 1/2

Let us consider a population in which every individual is classified according to two criteria: It is or it is not HIV carrier and it belongs or does not belongs to a risk group that we will call R. The probability chart for this study is:

 Carrier (A) Not carrier (Ac ) Belongs to R (B) 0.003 0.017 0.020 Does not belong to (Bc ) 0.003 0.977 0.980 0.006 0.994 1.000

In this population, the probability for an individual to be a carrier is P(A) = 0.006 and the probability for it to be a carrier and belong to the risk group R is P(A ∩ B) = 0.003.

Given that a person, selected at random, belongs to the R group, what is the probability that it is a carrier?

p ( A | B ) = p ( A B ) / p (B) = [0.003] / [0.020] = 0.150

Which means that 150 of the 1000 individuals in the risk group R are “probably” HIV carriers.

Let us calculate the probability of a person being HIV carrier, given that they do not belong to the risk group R.

p ( A | BC ) = p ( A BC ) / p (BC) = [0.003] / [0.980] = 0.00306

Meaning that only 3 of every 1000 individuals who do not belong to the risk group R, are “possible” HIV carriers.

### Example 3.

A woman is a carrier of the Duchenne sickness, which is the probability that her next son has this sickness?

According to Mendel’s laws, every possible genotypes for the son of a carrier mother (xX) and a normal father (XY) are xX, xY, XX, XY and have the same probability. The sample space is W= {xX, xY, XX, XY}. The event A = {sick son} corresponds to the genotype xY, it has, therefor, according to the classical definition of probability:
p(A) = 1/4 = 0,25

The woman has offspring and it is male, what is the chance that he has the sickness?

We define the event B = {being male} = {xY, XY}
The probability we want to know is p ( A | B )
applying the definition for B
p(B) = 0,5; A ⊂ B = {xY}; p (A⊂ B) = 0,25; p ( A | B ) = 0,25/0,5 = 0,5

If we know that he is male, the sample space has changed, now it is B. Therefor, we can calculate p(A|B) applying the classical definition of probability to the new sample space:
p(A|B) = 1/2 = 0,5

### Example 4.

We know that 50% of the population smokers and that 10% smokers and has high blood pressure. Which is the probability that a smoker has high blood pressure?

A={having high blood pressure} , B ={being smoker}
A⊂B = {having high blood pressure and being smoker}
p(A|B) = 0,10/0,50 = 0,20

The previous formula can be written as p(A⊂ B) = p(B) p(A|B) = p(A) p(B|A)

This is called the multiplication rule, which can be generalized to more events:
p(A
1A2A3) = p((A1A2)A3) = p(A1A2) p(A3|A1A2) =

p(A1) p(A2|A1) p(A3|A1A2)

In general p(A1A2A3…) = p(A1) p(A2|A1) p(A3|A1A2) … called principle of the compound probabilities and is specially useful for those situations en which conditioned probabilities are easier to obtain than intersection probabilities.

### Example 5.

It is known from previous studies that 0,1% of the population has vascular problems. A study on individuals with vascular problems reveals that 20% of them are atheroma plaques. If 10% of the individuals with atheroma plaques is exposed to sudden death by thrombus detachment, what probability does an individual, any individual, of being exposed to sudden death by thrombus detachment of an atheroma plaque?

A1= {vascular problems}; A2= {atheroma plaques}; A3= {exposure to sudden death by ….}
p(A
1) = 0,001; p(A2|A1) = 0,20; p(A3|A1 A2) = 0,1
p(A1
A2A3) = 0,001 x 0,20 x 0,1 = 0,000002