Question

If E₁ and E₂ are equally likely, mutually exclusive and exhaustive events and P(A/E₁) = 0.2, P(A/E₂) = 0.3. Find P(E₁/A).

Answer 1

Answer:

0.4

Step-by-step explanation:

Hi,

Given that  E₁ and E₂ are equally likely, mutually exclusive and

exhaustive events,

P(E₁) = P(E₂) Since they are equally likely

P(E₁ ∩ E₂) = 0, since they are mutually exclusive

P(E₁ ∪ E₂) = 1 , since they are exhaustive

But P(E₁ ∪ E₂) = P(E₁) + P(E₂) - P(E₁ ∩ E₂)

⇒ P(E₁) = P(E₂) = 1/2

P(E₁ ∩ A) = P(E₁)*P(A/E₁)

= 1/2*0.2 = 0.1

P(E₂ ∩ A) = P(E₂)*P(A/E₂)

= 1/2*0.3 = 0.15

P(A) = P(E₁ ∩ A) + P(E₂ ∩ A)

= 0.1 + 0.15

= 0.25

P(E₁/A) = P(E₁ ∩ A)/P(A)

= 0.1 / 0.25

= 0.4

Hope, it helps !