Step-by-step explanation:
The prize is equally likely to have been placed behind doors 1, 2, and 3, so the probabilities of D₁, D₂, and D₃ are equal.
P(D₁) = P(D₂) = P(D₃) = 1/3
The contestant initially chooses door 1, so the host will not open it.
P(H₁|D₁) = P(H₁|D₂) = P(H₁|D₃) = 0
The host will not open the door with the prize.
P(H₂|D₂) = P(H₃|D₃) = 0
The remaining probabilities for each door are equal.
P(H₂|D₁) = P(H₃|D₁) = 1/2
P(H₃|D₂) = 1
P(H₂|D₃) = 1
