Question

A certain medical test is known to detect 72% 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

Solution:
Given in the question
P(people who are afflicted with the disease Y) = 0.72
Number of sample = 10
Here we will use binomial distribution probability because all events are independent to each other
P(X=n |N,p) = NCn*(p^n)*(1-p)^(N-n)
Solution(a)
We need to calculate the probability that the test will show All 10 have the disease i.e. P(X=10) = ?
P(X=10 | 10,0.72) = 10C10*(0.72)^10 *(1-0.72)^(10-10) = 1*0.0374*1 = 0.0374
So there is 3.74% probability that the test will show All 10 have the disease.
Solution(b)
We need to calculate the probability that the test will show At least 8 have the disease i.e. P(X>=8) = ?
P(X>=8) = P(X=8) + P(X=9) + P(X=10) = 10C8*(0.72)^8 *(1-0.72)^(10-8) +10C9*(0.72)^9 *(1-0.72)^(10-9) +10C10*(0.72)^10 *(1-0.72)^(10-10) = 0.2548 + 0.1456 + 0.0374 = 0.4378
So there is 43.78% probability that the test will show At least 8 have the disease.
Solution(c)
We need to calculate the probability that the test will show At Most 4 have the disease i.e. P(X<=4) = ?
P(X<=4) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) = 10C0*(0.72)^0 *(1-0.72)^(10-0) +10C1*(0.72)^1 *(1-0.72)^(10-1) +10C2*(0.72)^2 *(1-0.72)^(10-2)+ 10C3*(0.72)^3 *(1-0.72)^(10-3) +10C4*(0.72)^4 *(1-0.72)^(10-4) = 0.0000003 + 0.00008 + 0.0009 + 0.0060 + 0.0272 = 0.0342
So there is 3.42% probability that the test will show At most 4 have disease.