The sample space of an experiment of tossing three coins is S = {TTT, HHH, TTH, HHT, THT, HTH, HTT, THH}. Since the sample space comprises of 8 elements, therefore, the probability of occurring each sample point is â…›. Let A and B be the events of displaying 2 heads and 1st coin showing tail respectively. Then, A = {HHT, HHH, THH, HTH} and B = {THT, THH, TTT, TTH}. Therefore P(A) = P({HHT}) + P ({HHH}) + P ({THH}) + P ({HTH}) = â…› + â…› + â…› + â…› = ½. Similarly, P(B) = P({THT}) + P({THH}) + P ({TTT}) + P ({TTH}) = = â…› + â…› + â…› + â…› = ½. Also, A ∩ B = {THH} and P({THH}) = P(A ∩ B) = â…›.

The sample point of B which is favorable to event A is THH. Thus, P(A) considering B as the sample space (S) = ¼. This P(A) is known as the conditional probability of A provided B has already occurred. The conditional probability of an event is denoted by P (A|B). Thus, from the above case P(A|B) = ¼.

Or, \(P(A|B) = \frac{Total\;events\;favourable \;to \;A\cap B }{Total\;events \;favourable \;to \;B}\)

Now, dividing both numerator and denominator by total elementary events of the S (sample space) we get:

\(P(A|B) = \frac{\frac{n(A\cap B)}{nS}}{\frac{n(B)}{nS}}=\frac{P(A\cap B)}{P(F)}\)The above equation is valid only when P(A) ≠ 0. Hence, the conditional probability can be described as:

- 0 ≤ P (P|Q) ≤ 1
- P (P′|Q) = 1 – P (P|Q)
- P ((P ∪ Q)|R) = P (P|R) + P (Q|R) – P ((P ∩ Q)|R)
- P (P ∩ Q) = P (P) P (Q|P), P (P) ≠ 0
- P (P ∩ Q) = P (P) P (P|Q), P (Q) ≠ 0

If P and Q are independent, then

- P (P ∩ Q) = P (P) P (Q)
- P (P|Q) = P (P), P (Q) ≠ 0
- P (Q|P) = P (Q), P(P) ≠ 0

### Probability Class 12 Practice Questions

- P and Q are two events such that P(P) ≠ 0. Find P(Q|P), if (i) P is a subset of Q (ii) P ∩ Q = φ
- If a coin is tossed 9 times, find the probability of
- exactly 7 heads
- at least 6 heads
- at most 7 head

- A man speaks truth 4 out of 5 times. He throws a die and reports that it is a 3. Determine the probability that the number is actually 3.
- Let P and Q be independent events with P(P) = 0.3 and P(Q) = 0.4. Find P(P ∩ Q), P(P ∪ Q), P (P|Q), P (Q|P).

**Also Read**

Bays Theorem | Probability Distribution of Random Variable |

Total Probability Theorem | Multiplication Rule of Probability |