Difference between revisions of "2021 ZeMC10 Problems/Problem 3"

(Created page with "==Problem== A treatment has a <math>90</math> percent success rate of curing those with CUBIC-27. If three patients with CUBIC-27 are selected at random, what is the probabili...")
 
(Problem)
 
Line 1: Line 1:
 
==Problem==
 
==Problem==
A treatment has a <math>90</math> 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 succesfully cured?
+
A treatment has a <math>90</math> 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?
  
 
<math>\textbf{(A) } 0.9 \% \qquad \textbf{(B) } 2.7 \% \qquad\textbf{(C) }  8.1 \% \qquad\textbf{(D) } 72.9 \% \qquad\textbf{(E) } 99.9 \%</math>
 
<math>\textbf{(A) } 0.9 \% \qquad \textbf{(B) } 2.7 \% \qquad\textbf{(C) }  8.1 \% \qquad\textbf{(D) } 72.9 \% \qquad\textbf{(E) } 99.9 \%</math>
 +
 +
==Solution==
 +
 +
 +
The first person of being cured successfully is <math>\frac{9}{10}</math>. The second person of being FAILED to be cured is <math>\frac{1}{10}</math>. The third person of being FAILED to be cured is <math>\frac{1}{10}</math>.
 +
 +
<math>\frac{9}{10}*\frac{1}{10}*\frac{1}{10}=\frac{9}{1000}</math>
 +
 +
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.
 +
 +
<math>\frac{9}{1000} * 3 = \frac{27}{1000}</math>
 +
 +
Then answer is (B)

Latest revision as of 19:50, 1 September 2022

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)