Question

A certain medical test is known to detect 56% of the people who are afflicted with the disease Y. If 10 people with the disease are administered the test, what is the probability that the test will show that: All 10 have the disease, rounded to four decimal places? At least 8 have the disease, rounded to four decimal places? At most 4 have the disease, rounded to four decimal places?

Answer 1

Answer:

The probability that all 10 have the disease is 0.0030

The probability that at least 8 have the disease is 0.1111

The probability that at most 4 have the disease is 0.2407

Step-by-step explanation:

Consider the provided information.

A certain medical test is known to detect 56% of the people who are afflicted with the disease Y. i.e p = 56% = 0.56

Let X be the number number of people who are afflicted with the disease Y.

10 people with the disease are administered the test. n=10

As we know p = 0.56 so q = 1-0.56 = 0.44

Part (A) All 10 have the disease, rounded to four decimal places.

$\binom{n}{k}p^kq^{n-k}$

Substitute respective values in the above formula.

$\binom{10}{10}(0.56)^{10}(0.44)^{10-10}=0.0030$

The probability that all 10 have the disease is 0.0030

Part (B) At least 8 have the disease.

P(X≥8)=P(X=8)+P(X=9)+P(X=10)

$P(X\geq 8)=\binom{10}{8}(0.56)^{8}(0.44)^{2}+\binom{10}{9}(0.56)^{9}(0.44)^{1}+\binom{10}{9}(0.56)^{10}(0.44)^{0}$

$P(X\geq 8)=0.11112$

The probability that at least 8 have the disease is 0.1111

Part (C) At most 4 have the disease,  

P(X≤4)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)

$P(X\leq 4)=\binom{10}{0}(0.56)^{0}(0.44)^{10}+\binom{10}{1}(0.56)^{1}(0.44)^{9}+\binom{10}{2}(0.56)^{2}(0.44)^{8}+\binom{10}{3}(0.56)^{3}(0.44)^{7}+\binom{10}{4}(0.56)^{4}(0.44)^{6}$

$P(X\leq 4)=0.2407$

The probability that at most 4 have the disease is 0.2407