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Conditional Probability

 

Table of Content


Independent Events

The events which are not affected by another events are known as Independent Events.

Two events A and B are independent if event A does not affect the occurrence of event B

Example: Rolling a dice

Rolling a dice

Fig 1: Rolling a dice

The event of getting a 6 the first time a die is rolled and event of getting a 6 the second time is completely independent and isolated.


Dependent Events

The events which are dependent on other events and they can be affected by previous events.

Example:

Fig 2: Marbles in a bag

2 blue and 3 red marbles are there in a bag

The chances of getting blue marbles is 2/5. But after one marble is taken out, the chances change:

  • If we got red marble in the previous event, the chance of blue marble in the next is 2/4

  • If we got blue marble in the previous event, then the chance of blue marble in the next is ¼
     

What is Conditional Probability?

Conditional Probability simply measures “How to measure or handle Dependent Events

In probability theory, Conditional Probability is a measure of the probability of an event given that (by assumption, presumption or evidence) another event has occurred. If the event A is unknown and the event B is known or assumed to have occurred, "the conditional probability of A given B", or "the probability of A under the condition B", is usually written as P (A|B), or sometimes PB (A).

Example, the probability that any person has a cough on a day may be only 5%. But if we know or assume that the person has a cold, then they are much more likely to be coughing. The conditional probability of coughing given that you have a cold might be a much higher 75%.

Conditional Probability

Fig 3: Conditional Probability

How to calculate Conditional Probability?

Let S be a finite sample space, associated with the given random experiment, containing equally likely outcomes. Then we have the following result.

Statement: Conditional probability of event A given that event B has already occurred is given by

Fig 4: Venn Diagram – Conditional Probability

P(A/B) = P(A∩B) / P(B) ,  P(B) ≠0

Let S be a sample space associated with the given random experiment and n(S) be the number of sample points in the sample space S. Since it is given that event B has already occurred, therefore our sample space reduces to event B only, which contains n(B) sample points. Event B is also called reduced or truncated sample space. Now out of n(B) sample points, only n(A∩B) sample points are favourable for the occurrence of event A. Therefore by definition of probability.

P(A/B) = n (A∩B) / n(B),                                    n(B)≠0

             = n(A∩B)/ n(S) / n(B)/n(S)

Therefore, P(A/B) = P(A∩B)/P(B)                      P(B)≠0

Similarly P(B/A) = P(A∩B)/ P(A)                         P(A)≠0


Law of Total Probability

Consider a country with three provinces B1, B2 and B3 (i.e country is partitioned into 3 disjoint sets B1, B2 and B3). To calculate the total forest area knowing the area of provinces B1, B2 and B3 as 100Km2, 50 Km2 and 150 Km2. obviously, we will add all the three provinces areas to get total area of forest.

100Km2 +50 Km2 + 150 Km2 = 300 Km2

The same is the idea behind law of total probability. In which the area of the forest is replaced by the probability of an event A. In a case where the partition is B and B’ , then for any 2 events A and B.

P(A)= P( A ∩ B) + P( A ∩ B’)

By Conditional Probability, P ( A ∩ B ) = P(A | B) P(B)

Therefore, P (A) = P (A | B) P( B) + P(A | B’) P(B’)

Hence, we can state a general version of the formula for Sample Space S as below:

If B1, B2, B3,⋯  is a partition of the sample space S, then for any event A we have

P(A)=∑ P(A ∩ Bi) = ∑ P(A | Bi) P(Bi)

As it can be seen from the figure, A1, A2, A3 , A4 and A5 form a partition of the set A, and thus by the third axiom of probability

P(A)=P(A1)+P(A2)+P(A3)+P(A4)+P(A5).
 

Examples of Conditional Probability


Q1. Find the probability that a single toss of die will result in a number less than 4 if it is given that the toss resulted in an odd number.

Sol. Let event A: toss resulted in an odd number and event B: number is less than 4

Therefore, A= {1, 3, 5}

P (A) =3/6 = ½

B= {1, 2, 3}

A∩B = {1, 3}

P (A∩B) =2/6=1/3

Therefore, required probability = (P number is less than 4 given that it is odd)

 = P (B/A) = P (A∩B)/P (A) =1/3 / ½ =2/3
 

Q2. The probability that it is Friday and a student is absent is 0.03. Since there are 5 school days in week, the probability that it is Friday is 0.2. What is the probability that a student is absent given that today is Friday.

Sol.


Q3.  A jar contains black and white marbles. Two marbles are chosen without replacement. The probability of selecting a black marble and then a white marble is 0.34, and the probability of selecting a black marble on the first draw is 0.47. What is the probability of selecting a white marble on the second draw, given that the first marble was drawn black? 

Sol.


Q4. If P (A’) =0.7, P (B) = 0.7, P(B/A)=0.5, find P(A/B) and P(AUB).

SolSince P (A’) =0.7,

P (A) =1-P (A’) =1-0.7=0.3

Now P (B/A) = P (A∩B) / P (A)

0.5= P (A∩B) / 0.3

P (A∩B) =0.15

Again, P (A/B) = P (A∩B) /P (B)

=0.15 / 0.7

P (A/B) = 3/14

Further, by addition theorem

P (AUB) = P (A) +P (B) – P (A∩B)

             =0.3+0.7-0.15

             =0.85

 

Frequently Asked Questions (FAQs)


Q1. Take the case of previous illustration in which a bag contains 3 Red, 2 Green and 4 Yellow balls. A ball is drawn and if it is Red or Yellow. it is replaced otherwise not. Find the probability that second ball

drawn is a yellow ball.

Sol. First of all, we should be very clear about the event of which probability is to be found and the events on which it depends

Let's A1: First ball is either red or yellow

A2: First ball is green

A: Second ball drawn is yellow

We have to calculate P(A) which definitely depends on A1 and A2.

Therefore, P(A) = P(A/A1) P(A1) + P(A/A2) P(A2

P(A/A2) = Probability of A when A1 is already happen.

= 4/9 (i.e. probability of A given A1)

P(A1) = 7/9

P(A/A2) = Probability of A when A2 is already happened

= (Probability of A given A2)

= 4/8

P(A2) = 2/9

Hence, P(A) = 4/9×7/9+4/8×2/9 = 37/81
 

Q2.  At ABC Senior Secondary school, the probability that a student takes Science and English is 0.087. The probability that a student takes Science is 0.68. What is the probability that a student takes English

given that the student is taking Science?

Sol. We need to find the probability of English given Science i.e.

P (English | Science) = P (Science and English) / P (Science) = 0.087 / 0.68 = 0.13 = 13%.
 

Q3. A jar contains red and green toys. Two toys are drawn without replacement. The probability of selecting a red toy and then a green toy  is 0.34, and the probability of selecting a red toy on the first draw is 0.47. What is the probability of selecting a green toy in the second draw, given that the first toy drawn was red?

Sol. Clearly it is a question of conditional probability. We need to find the probability of drawing a green toy having known that the first toy drawn was red.

P (green | red) = P (red and green) / P (red) = 0.34 / 0.47 = 0.72 = 72%.


Q4. At ABC Senior Secondary school, the probability that a student takes Science and English is 0.087. The probability that a student takes Science is 0.68. What is the probability that a student takes English given that the student is taking Science?

Sol. We need to find the probability of English given Science i.e.

P (English | Science) = P (Science and English) / P (Science) = 0.087 / 0.68 = 0.13 = 13%.

 

Related Resources

http://www.askiitians.com/iit-jee-algebra/probability/

http://www.askiitians.com/iit-papers/iit-jee-papers.aspx

http://www.askiitians.com/school-exams/cbse/class-12/past-papers.html

http://www.askiitians.com/iit-papers/iit-jee-papers.aspx
 

Watch this Video for more reference
 


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