Conditional Probability Examples and Properties

Let A and B be two events with a random experiment then there are two possibilities. Remember, this is really important for NEET exam and IIT JAM exam aspirants :

CASE I: Occurrence or non-occurrence of one event affects the probability of occurrence or non-occurrence of another event.

CASE II: Occurrence or non-occurrence of one event does not affect the probability of occurrence or non-occurrence of another event.

Let us consider the case(I).

Let three coins be tossed. The sample space for the experiment is

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Let A and B be two events defined as:

A: Tail on the first coin, A = { THH,THT,TTH,TTT}

B: At least two heads, B = {HHH,HHT,HTH,THH}

We have P(A) =4/8 = 1/2 and P(B) =  4/8 = 1/2 

A ∩ B = {THH}

= P(A ∩B) = 1/8

Now, suppose we are given that the first coin will show a tail, i.e., event A has already occurred. Then what will be the probability of the occurrence of event B? Since event A has already occurred, so all such cases where the first coin does not show tail cannot be considered while finding the probability of B. This condition reduces the sample space from S to its subset A. Now in this new sample space the outcomes/elements favourable to B is {THH}.

Thus, probability of B considering A as the sample space =  1/4

Or probability of B given that A has already occurred = 1/4.

This probability of the event B is called the conditional probability of B given that A has already occurred.

P(B/A) denotes the probability of the occurrence of B when A has already occurred.

Similarly, P(A/B) denotes the probability of occurrence of A when B has already occurred.

Note that the elements of B which favour the event A are the common elements of A and B, i.e., the sample points of A∩B. Thus, we can express the conditional probability of A when B has already occurred as:

$$\frac{\text{P(A/B) = Number of elementary events favourable to}A\cap B}{\text{Number of elementary events favourable to}B}=\frac{n(A\cap B)}{n\left(B\right)}$$

Dividing both numerator and denominator by total number of elementary events of the sample space. Then P(A/B) can be written as

$$P(A/B)=\frac{\frac{n(A\cap B)}{n\left(S\right)}}{\frac{n\left(B\right)}{n\left(S\right)}}=\frac{P(A\cap B)}{P\left(B\right)}$$ Thus, we can say,

If A and B are two events associated with the same sample space of the random experiment, the conditional probability of the event A when B has already occurred is given by

$$P(A/B)=\frac{P(A\cap B)}{P\left(B\right)},\text{ if }P\left(B\right)\ne 0.$$

Let us take a look at the properties of conditional properties that are useful for IISER exam and CUET exam aspirants:

1. Non-negativity:  For any events A and B, the conditional probability P(A|B) is always greater than or equal to 0. This aligns with the basic axioms of probability, where probabilities cannot be negative.
2. Normalisation: If we consider the conditional probability of the sample space (S) given event B, or the conditional probability of event B given event B, it will always be equal to 1. This means that if we know B has occurred, then the probability of something within the sample space happening is certain.

Mathematically: P(S|B) = P(B|B) = 1.

3. Additivity: If we have mutually exclusive events (A1, A2, A3, ...), meaning they cannot occur at the same time, then the conditional probability of their union (occurring given B) is the sum of their individual conditional probabilities. 

Mathematically: P (A1 ∪ A2 ∪ A3 ... | B) = P(A1|B) + P(A2|B) + P(A3|B) + ... 

4. Multiplication Rule (and Chain Rule): The multiplication rule states that the probability of both A and B occurring is the product of the probability of A given B and the probability of B.

Mathematically: P (A ∩ B) = P(A|B)  P(B).

This can be extended to multiple events, forming the chain rule.

5. Relationship to Independent Events:
   •  If events A and B are independent, meaning the occurrence of one does not affect the probability of the other, then P(A|B) = P(A) and P(B|A) = P(B).
   •  In simpler terms, if events are independent, knowing that one has happened does not change the likelihood of the other happening.

1. Find the value of P(A|B) if P(A) = ⅚ and P(B) = 7/6 and P (A ∩ B) = ⅙ 

Solution. P(A|B) = P(A ∩ B)/P(B) = ⅙ / 7/6 = 1/7 

2. A dice is thrown twice, and the numbers appearing are observed. If the sum of the numbers appearing on the dice is 5, what is the probability of number 2 appearing at least once?

Solution. Suppose that F is the event when the sum of the numbers is 5 

F = { (1,4), (2,3), (3,2), (4,1)} 

Therefore, P(F) = 4/36 = 1/9

Now, let E be the event that the number 2 appears at least once. 

E = { (1,2), (2,2), (3,2), (4,2), (5,2), (6,2), (2,1), (2,3), (2,4), (2,5), (2,6)} 

P(E) = 11/36

E ∩ F = {(2.3), (3.2)}

P(E ∩ F) = 2/36 = 1/18 

Therefore, 

P(E|F) = P(E ∩ F)/P(F) = 1/18 /  1/9 = ½.