2021 ZeMC10 Problems/Problem 3

Problem

A treatment has a $90$ percent success rate of curing those with CUBIC-27. If three patients with CUBIC-27 are selected at random, what is the probability that exactly one of the three patients will be successfully cured?

$\textbf{(A) } 0.9 \% \qquad \textbf{(B) } 2.7 \% \qquad\textbf{(C) } 8.1 \% \qquad\textbf{(D) } 72.9 \% \qquad\textbf{(E) } 99.9 \%$

Solution

The first person of being cured successfully is $\frac{9}{10}$ . The second person of being FAILED to be cured is $\frac{1}{10}$ . The third person of being FAILED to be cured is $\frac{1}{10}$ .

$\frac{9}{10}*\frac{1}{10}*\frac{1}{10}=\frac{9}{1000}$

BUT, there are 3 possiblilities, one being the first getting cured, one being the second person being cured, and the last is the third person being cured.

$\frac{9}{1000} * 3 = \frac{27}{1000}$

Then answer is (B)

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2021 ZeMC10 Problems/Problem 3

Problem

Solution