Question

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?

Answer 1

The final answer is approximately
`\(0.0524\).`

`\[ P(X = 0) = \binom{15}{0} \cdot 0.24^0 \cdot (1 - 0.24)^(15) \]\\`

`\[ P(X = 0) = 1 \cdot 1 \cdot (0.76)^(15) \]\[ P(X = 0) \approx 0.0021 \]\[ P(X = 1) = \binom{15}{1} \cdot 0.24^1 \cdot (1 - 0.24)^(14) \]\[ P(X = 1) = 15 \cdot 0.24 \cdot (0.76)^(14) \]\[ P(X = 1) \approx 0.0132 \]\[ P(X = 2) = \binom{15}{2} \cdot 0.24^2 \cdot (1 - 0.24)^(13) \]\[ P(X = 2) = 105 \cdot 0.24^2 \cdot (0.76)^(13) \]\[ P(X = 2) \approx 0.0371 \]`

Now, add these probabilities:


`\[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \]\[ P(X \leq 2) \approx 0.0021 + 0.0132 + 0.0371 \]\[ P(X \leq 2) \approx 0.0524 \]`