 # Probability Class 12 Notes - Chapter 13

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 ⅛. 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})  = ⅛ + ⅛ + ⅛ + ⅛ = ½. Similarly, P(B) = P({THT}) + P({THH}) + P ({TTT}) + P ({TTH}) = = ⅛ + ⅛ + ⅛ + ⅛ = ½. Also, A ∩ B = {THH} and P({THH}) = P(A ∩ B) = ⅛.

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

1. 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
2. If a coin is tossed 9 times, find the probability of