Question

A coin is tossed twice. If E and F denote occurrence of head on first toss and second toss respectively, then what is P(E ∪ F) equal to?
1. \(\dfrac{1}{4}\)
2. \(\dfrac{1}{2}\)
3. \(\dfrac{3}{4}\)
4. \(\dfrac{1}{3}\)

Answer 1

Correct Answer - Option 3 : \(\dfrac{3}{4}\)

Concept:

For the independent events A and B, 

• P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
• P(A ∩ B) = P(A) × P(B)

 

Calculation:

Probability of getting head on first toss P(E) = \(1\over2\)

Probability of getting head on first toss P(F) = \(1\over2\)

Since the events are independent P(E ∩ F) = P(E) × P(F)

P(E ∩ F) = \(1\over2\) × \(1\over2\) = \(1\over4\)

P(E ∪ F) = P(E) + P(F) - P(E ∩ F)

P(E ∪ F) = \(1\over2\) + \(1\over2\) - \(1\over4\)

P(E ∪ F) = \(3\over4\)