This is a binomial probability problem where:

• n = 7 (number of trials/days)
• k = 3 (number of successes/sunny days)
• p = 0.9 (probability of success/sunny day)
• q = 1 - p = 0.1 (probability of failure/not sunny day)

The binomial probability formula is: P(k) = C(n,k) * p^k * q^(n-k) Where C(n,k) = n!/((n-k)!*k!)

Calculation: C(7,3) = 7!/(4!*3!) = (7×6×5)/(3×2×1) = 35 p^k = 0.9^3 = 0.729 q^(n-k) = 0.1^4 = 0.0001

P(3) = 35 × 0.729 × 0.0001 = 35 × 0.0000729 = 0.0025515 ≈ 0.00255

This matches the calculation provided in the text: P(3) = 7!/((7-3)!*3!) * 0.9³ * (1-0.9)⁷⁻³ = 0.00255