Question 2

**Setting up (assuming the students solve independently):** Let p₁, p₂, p₃ be the probabilities that S₁, S₂, S₃ can solve the problem. - P(U) = 1 − (1−p₁)(1−p₂)(1−p₃) = 1/2 ⇒ (1−p₁)(1−p₂)(1−p₃) = 1/2 - P(V) = P(S₁ | S₂ᶜ ∩ S₃ᶜ) = p₁ = 1/10 (by independence) - P(W) = p₂(1−p₃) = 1/12 From the first equation with p₁ = 1/10: (9/10)(1−p₂)(1−p₃) = 1/2 ⇒ (1−p₂)(1−p₃) = 5/9 Adding this to P(W): (1−p₂)(1−p₃) + p₂(1−p₃) = (1−p₃) = 5/9 + 1/12 = 20/36 + 3/36 = 23/36 So p₃ = 1 − 23/36 = 13/36. Hence P(T) = 13/36.

No written reasoning was returned for this item.