A new drug on the market is known to cure 24% of patients with colon cancer. If a group of 15 patients is randomly selected, what is the probability of observing, at most, two patients who will be cured of colon cancer? 15 choose 2 (0.24) 2 (0.76) 13 15 choose 0 (0.76) 15 + 15 choose 1 (0.24) 1 (0.76) 14 + 15 choose 2 (0.24) 2 (0.76) 13 1 - 15 choose 2 (0.24) 2 (0.76) 13 15 choose 0 (0.76) 15 1 - 15 choose 0 (0.76) 15

P = ₁₅C₀ (0.76) ¹⁵ + ₁₅C₁ (0.24) ¹ (0.76) ¹⁴ + ₁₅C₂ (0.24) ² (0.76) ¹³

Step-by-step explanation:

Binomial probability:

P = nCr pʳ qⁿ⁻ʳ

where n is the number of trials,

r is the number of successes,

p is the probability of success,

and q is the probability of failure (1-p).

Here, n = 15, p = 0.24, and q = 0.76.

We want to find the probability when r is at most 2, which means r = 0, r = 1, and r = 2.

P = ₁₅C₀ (0.24) ⁰ (0.76) ¹⁵⁻⁰ + ₁₅C₁ (0.24) ¹ (0.76) ¹⁵⁻¹ + ₁₅C₂ (0.24) ² (0.76) ¹⁵⁻²

P = ₁₅C₀ (0.76) ¹⁵ + ₁₅C₁ (0.24) ¹ (0.76) ¹⁴ + ₁₅C₂ (0.24) ² (0.76) ¹³