Conditional Probability for Grade XII

The probability of an event occurring when another event has already occurred is known as conditional probability. Also play conditional probability worksheets to improve your math skills.

The probability that event B occurs, given that event A has already occurred is

   P(B|A) = P(A and B) / P(A).

This formula is derived from the general multiplication principle and algebra.

Since, we know that event A has occurred, we conclude that the sample space has reduced. The sample space S, is reduced to a sample space of A since we know A has taken place. The number in the event divided by the number in the sample space continues. The value of both A and B should be in A since A has already taken place and divided by the number in A. Dividing numerator and denominator of the right hand side by the number in the sample space S, we have the probability of A and B divided by the probability of A.

Conditional Probability

Suppose we put a distribution function to a sample space and then learn that an event E has occurred.

Problem arise that how can 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 jE).

Example 1. Lets take an example of a die rolled once. Let X be the outcome.

Let F be the event fX = 6g, and let E be the event fX > 4g. We assign the

distribution function m(!) = 1=6 for ! = 1; 2; : : : ; 6. Thus, P (F ) = 1=6. Now

suppose that the die is rolled and we are told that the event E has occurred. This

leaves only two possible outcomes: 5 and 6. In the absence of any other information,

we would still regard these outcomes to be equal likely, so the probability of F

becomes 1/2, making P (F jE) = 1=2. 2

Example 2. Let us assume that, Out 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. 2.

Example 1:

The question, “ Do you know maths?" was asked to 100 people.

.	Yes	No	Total
Male	25	14	39
Female	24	37	61
Total	49	51	100

• What is the probability of a randomly selected person who is female and knows math? This is a joint probability. The number of "Female and Knows Math" divided by the total = 24/100 = 0.24
• What is the probability of a randomly selected person who is a male excluding whether he knows math or does not know the same? It can be calculated by taking the total for male divided by the total =39/100 = 0.39. Since the trait of knowing maths is not asked, male of each class can be mentioned.
• What is the probability of a randomly selected person knowing math? As gender is not mentioned this question becomes a marginal probability, the total who know maths divided by the total = 49/100 = 0.49.
• What is the probability of a randomly selected male knowing math? What is the probability of the males knowing math? So, now we have males knowing math of 39 males, so 25/39 = 0.641.
• What is the probability that a randomly selected person knowing maths is male? Now, we have a person knowing maths and asked to find the probability that the person knowing math is also male. There are 25 male who know math out of total 49 people knowing math, so 25/49 = 0.5104.

Example 2:

There are three major manufacturing companies that make a product: Aberations, Brochmailians, and Chompielians. Aberations has a 50% market share, and Brochmailians has a 30% market share. 5% of Aberations' product is defective, 7% of Brochmailians' product is defective, and 10% of Chompieliens' product is defective.

This information can be placed into a joint probability distribution

Company	Good	Defective	Total
Aberations	0.50-0.025 = 0.475	0.05(0.50) = 0.025	0.50
Brochmailians	0.30-0.021 = 0.279	0.07(0.30) = 0.021	0.30
Chompieliens	0.20-0.020 = 0.180	0.10(0.20) = 0.020	0.20
Total	0.934	0.066	1.00

The percent of the market share for Chompieliens wasn't given, but since the marginals must add to be 1.00, they have a 20% market share.

Notice that the 5%, 7%, and 10% defective rates don't go into the table directly. This is because they are conditional probabilities and the table is a joint probability table. These defective probabilities are conditional upon which company was given. That is, the 7% is not P(Defective), but P(Defective|Brochmailians). The joint probability P(Defective and Brochmailians) = P(Defective|Brochmailians) * P(Brochmailians).

The "good" probabilities can be found by subtraction as shown above, or by multiplication using conditional probabilities. If 7% of Brochmailians' product is defective, then 93% is good. 0.93(0.30)=0.279.

• What is the probability a randomly selected product is defective? P(Defective) = 0.066
• What is the probability that a defective product came from Brochmailians? P(Brochmailian|Defective) = P(Brochmailian and Defective) / P(Defective) = 0.021/0.066 = 7/22 = 0.318 (approx).
• Are these events independent? No. If they were, then P(Brochmailians|Defective)=0.318 would have to equal the P(Brochmailians)=0.30, but it doesn't. Also, the P(Aberations and Defective)=0.025 would have to be P(Aberations)*P(Defective) = 0.50*0.066=0.033, and it doesn't.

The second question asked above is a Bayes' problem. Again, my point is, you don't have to know Bayes formula just to work a Bayes' problem.

Bayes' Theorem

Bayes' formula can be well defined by the assumption given further.

Let's say, each event is taken as to A, B, C, and D.

Bayes' formula is used to find the reverse conditional probability P(B|D). And P(D|B) is not a Bayes problem as discussed in the given problem.

It is assumed that D is made of three parts, the part of D in A, the part of D in B, and the part of D in C.

  

                                        P(B and D)                   
   P(B|D) =       P(A and D)  + P(B and D)  + P(C and D)

Inserting the multiplication rule for each of these joint probabilities gives

                                      P (D|B)*P(B)                        
   P(B|D) =         P(D|A)*P(A) + P(D|B)*P(B) + P(D|C)*P(C)

It can be well accepted that, it is much easier to take the joint probability divided by the marginal probability.

Bayes Probabilities

Our original tree measure gave us the probabilities for drawing a ball of a given color, given the urn chosen. We have just calculated the inverse probability that a

particular urn was chosen, given the color of the ball. Such an inverse probability is called a Bayes probability and may be obtained by a formula that we shall develop later. Bayes probabilities can also be obtained by simply constructing the tree measure for the two-stage experiment carried out in reverse order.

The paths through the reverse tree are in one-to-one correspondence with those in the forward tree, since they correspond to individual outcomes of the experiment, and so they are assigned the same probabilities. From the forward tree, we find that the probability of a black ball is

1 . 2 + 1. 1 = 9
2 5 2 2 20

The probabilities for the branches at the second level are found by simple division. For example, if x is the probability to be assigned to the top branch at the second level, we must have ,

9 . x = 1

20 5

or x = 4=9. Thus, P (IjB) = 4=9, in agreement with our previous calculations. The reverse tree then displays all of the inverse, or Bayes, probabilities.

Independent Events:

It often happens that the knowledge that a certain event E has occurred has no effect on the probability that some other event F has occurred, that is, that P (F jE) = P (F ). One would expect that in this case, the equation P (EjF ) = P (E) would also be true. In fact (see Exercise 1), each equation implies the other. If these

equations are true, we might say the F is independent of E. For example, you would not expect the knowledge of the outcome of the first toss of a coin to change

the probability that you would assign to the possible outcomes of the second toss, that is, you would not expect that the second toss depends on the first. This idea

is formalized in the following definition of independent events. Two events E and F are independent if both E and F have positive probability and if

P (EjF ) = P (E) ;

and

P (F jE) = P (F ) :

As noted above, if both P (E) and P (F ) are positive, then each of the above equations imply the other, so that to see whether two events are independent, only

one of these equations must be checked.

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