Bunuel wrote:
There are 5 treatments for a disease: A, B, C, D and E. If a doctor chooses 3 out of 5 for the treatment, what is the probability that A will be chosen?
A. 1/10
B. 1/5
C. 3/10
D. 3/5
E. 7/10
Total possible selection of any 3 out of 5 = 5C3 = 10
Total possible selection of 3 such that A is included = Selection of 2 out of 4 leaving A aside as selected = 4C2 = 6
Probability \(= \frac{6}{10} = \frac{3}{5}\)
Answer:Option D
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