Q.If A and B are two independent events such that P(A∪B)=0.6, P(A)=0.2, find P(B).
Concept understanding — Multiplication Theorem & Independence
Definition 12.15 (independence). Two events A,B are independent exactly when
P(A∩B)=P(A)⋅P(B).
When P(A),P(B)>0, this is exactly equivalent to P(B/A)=P(B) and P(A/B)=P(A) -- knowing A happened tells you nothing new about B's chances, and vice versa. This is the special case of the Multiplication Theorem where the conditional term collapses to the plain probability. Mutually independent events A1,…,An satisfy the stronger requirement that EVERY sub-collection multiplies out: P(Ai1∩⋯∩Aik)=P(Ai1)⋯P(Aik).
Theorem 12.8. If A,B are independent, then so are (i) Aˉ,B; (ii) A,Bˉ; (iii) Aˉ,Bˉ. (Proved via De Morgan's law: P(Aˉ∩Bˉ)=P(A∪B)=1−P(A∪B)=1−[P(A)+P(B)−P(A)P(B)]=[1−P(A)][1−P(B)]=P(Aˉ)P(Bˉ).)
Theorem 12.9. For events with P(A)=0, P(B)=0: (1) mutually exclusive events can NEVER be independent, and (2) independent events can NEVER be mutually exclusive. The reason is direct: independence needs P(A∩B)=P(A)P(B)>0 (a genuine positive overlap), while mutual exclusivity needs P(A∩B)=0 -- the two conditions contradict each other unless one event is essentially impossible.
Independence is a probability property, mutual exclusivity is a set-theoretic property. Whether two events are independent can only be checked from their PROBABILITIES (P(A∩B) vs P(A)P(B)); whether they are mutually exclusive is checked from the EVENTS themselves (A∩B=∅ or not). Typical applications: independent coin tosses/die rolls/card draws WITH replacement; a card draw WITHOUT replacement is generally NOT independent, since removing the first card changes the composition available for the second.
Use the independent-event form of the Addition Theorem, P(A∪B)=P(A)+P(B)−P(A)P(B).
P(B)=0.5.
Step 1. Independent-event Addition Theorem. Since A,B are independent, P(A∩B)=P(A)P(B), so P(A∪B)=P(A)+P(B)−P(A)P(B).
Step 2. Substitute known values. 0.6=0.2+P(B)−(0.2)P(B)=0.2+0.8P(B).
Step 3. Solve for P(B). 0.8P(B)=0.6−0.2=0.4 ⇒ P(B)=0.80.4=0.5.
P(B)=0.5.
Substitute P(A∩B)=P(A)P(B) into the Addition Theorem and solve the resulting linear equation for P(B).
- Using the plain (non-independent) Addition Theorem P(A∪B)=P(A)+P(B)−P(A∩B) with P(A∩B) left as an unknown, instead of substituting P(A)P(B) for it.
- CBSE 2022Set ANNUAL1 markMCQQ.If A and B are two events such that P(A)=0.4, P(B)=0.8 and P(B/A)=0.6, then P(A∩B) is:(a) 0.56(b) 0.96(c) 0.66(d) 0.24
›Reveal solutionSolution
P(A∩B)=P(B)−P(A∩B)=0.8−0.24=0.56.
First find P(A∩B) using the conditional probability formula: P(B/A)=P(A)P(A∩B), so P(A∩B)=P(B/A)⋅P(A)=0.6×0.4=0.24.
Since B splits into the part that overlaps with A and the part that doesn't, P(B)=P(A∩B)+P(A∩B).
So P(A∩B)=P(B)−P(A∩B)=0.8−0.24=0.56.
✓Final answerThe correct option is (a) 0.56.
- CBSE 2022Set ANNUAL1 markMCQQ.If X and Y be two events such that P(X/Y)=21, P(Y/X)=31 and P(X∩Y)=61, then P(X∪Y) is:(a) 61(b) 31(c) 32(d) 52
›Reveal solutionSolution
Solving for P(X)=21 and P(Y)=31 from the given conditionals, the addition rule gives P(X∪Y)=32.
From P(X/Y)=P(Y)P(X∩Y): 21=P(Y)1/6⇒P(Y)=1/21/6=31.
From P(Y/X)=P(X)P(X∩Y): 31=P(X)1/6⇒P(X)=1/31/6=21.
By the addition rule, P(X∪Y)=P(X)+P(Y)−P(X∩Y)=21+31−61=63+2−1=64=32.
✓Final answerThe correct option is (c) 32.
- CBSE 2019Set ANNUAL1 markMCQQ.It is given that the events A and B are such that P(A)=41, P(A/B)=21 and P(B/A)=32. Then P(B) is:(a) 32(b) 21(c) 61(d) 31
›Reveal solutionSolution
First get P(A∩B) from P(B∣A)=P(A∩B)/P(A), then use P(A∣B)=P(A∩B)/P(B) to solve for P(B)=1/3.
Given P(A)=41, P(A∣B)=21, P(B∣A)=32.
From P(B∣A)=P(A)P(A∩B): P(A∩B)=P(B∣A)⋅P(A)=32×41=61.
From P(A∣B)=P(B)P(A∩B): P(B)=P(A∣B)P(A∩B)=1/21/6=31.
✓Final answerThe correct option is (d) 31.
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