The prior probabilities for events A1, A2, and A3 are P(A1) = 0.50, P(A2) = 0.20, and P(A3) = 0.30. The conditional probabilities of event B given A1, A2, and A3 are P(B | A1) = 0.50, P(B | A2) = 0.40, and P(B | A3) = 0.20. (Assume that A1, A2, and A3 are mutually exclusive events whose union is the entire sample space.) Required:
Compute P(B â© A1), P(B â© A2), and P(B â© A3).

P( B∩ A₁ ) = 0.25

P( B∩ A₂ )  = 0.08

P( B∩ A₃  ) = 0.06

Step-by-step explanation:

As we know that

P( B∩ A₁ ) = P( B| A₁) × P(A₁)

P( B∩ A₂ ) = P( B| A₂) × P(A₂)

P( B∩ A₃ ) = P( B| A₃) × P(A₃)

As we have

P(A₁) = 0.50, P(A₂) = 0.20, and P(A₃) = 0.30.

P(B | A₁) = 0.50, P(B | A₂) = 0.40, and P(B | A₃) = 0.20

⇒P( B∩ A₁ ) = P( B| A₁) × P(A₁)

                   = 0.50(0.50)

                   = 0.25

⇒P( B∩ A₁ ) = 0.25

Now,

⇒P( B∩ A₂ ) = P( B| A₂) × P(A₂)

                   = 0.40(0.20)

                   = 0.08

⇒P( B∩ A₂ )  = 0.08

Now,

⇒P( B∩ A₃ ) = P( B| A₃) × P(A₃)

                   = 0.20(0.30)

                   = 0.06

⇒P( B∩ A₃  ) = 0.06

∴ we get

P( B∩ A₁ ) = 0.25

P( B∩ A₂ )  = 0.08

P( B∩ A₃  ) = 0.06

the solution is h=10 because you add 9 to each side of the equation.

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