Anew 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

The correct answer is When P = ₁₅C₀ (0.76)¹⁵ + ₁₅C₁ (0.24)¹ (0.76)¹⁴ + ₁₅C₂ (0.24)² (0.76)¹³

                     P = nCr pʳ qⁿ⁻ʳ

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

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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

it's

no one knows lol

step-by-step explanation: