Conditional Probability

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 P_B (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%.

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 100Km^2, 50 Km² and 150 Km^2. 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:

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).

Sol. Since 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

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

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

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

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

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