During the past week, a local medical clinic tested \(N\) individuals for two infections. If \(\dfrac13\) of those tested had infection \(A\) and, of those with infection \(A,\) \(\dfrac15\) also had infection \(B,\) how many individuals did not have both infections \(A\) and \(B?\)
A. \(\dfrac{N}{15}\)
B. \(\dfrac{4N}{15}\)
C. \(\dfrac{14N}{15}\)
D. \(\dfrac{N}5\)
E. \(\dfrac{4N}5\)
Answer: C
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During the past week, a local medical clinic tested \(N\) individuals for two infections. If \(\dfrac13\) of those teste
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deloitte247
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Total number of people tested for two infections = N
People with infection A = 1/3 of N = N/3
People with infection B and infection A = 1/5 of N/3 = N/15
Total no. of people who did not have both infection A and B
$$=\frac{N}{1}-\frac{N}{15}=\frac{15N-N}{15}=\frac{14N}{15}$$
Answer = C
People with infection A = 1/3 of N = N/3
People with infection B and infection A = 1/5 of N/3 = N/15
Total no. of people who did not have both infection A and B
$$=\frac{N}{1}-\frac{N}{15}=\frac{15N-N}{15}=\frac{14N}{15}$$
Answer = C
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Solution:M7MBA wrote: ↑Sun Nov 29, 2020 12:44 pmDuring the past week, a local medical clinic tested \(N\) individuals for two infections. If \(\dfrac13\) of those tested had infection \(A\) and, of those with infection \(A,\) \(\dfrac15\) also had infection \(B,\) how many individuals did not have both infections \(A\) and \(B?\)
A. \(\dfrac{N}{15}\)
B. \(\dfrac{4N}{15}\)
C. \(\dfrac{14N}{15}\)
D. \(\dfrac{N}5\)
E. \(\dfrac{4N}5\)
Answer: C
We know that N/3 individuals had infection A, and 1/5 of those (1/5 x N/3 = N/15) had both infections A and B. Therefore, N - N/15 = 14N/15 individuals do not have both infections.
Answer: C
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