Question

A certain medical test is known to detect 62% 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?

please show work~

Answer 1

Solution:- Given that p = 0.62, n = 10 q = 1-p = 1-0.62 = 0.38

a) P(X = 10) = 10C10*0.62^10*0.38^0 = 0.0084

b) P(X >= 8) = P(8) + P(9) + P(10) = 0.1419+0.0514+0.0084 = 0.2017

=> P(8) = 10C8*0.62^8*0.38^2 = 0.1419

=> P(9) = 10C9*0.62^9*0.38^1 = 0.0514

=> P(10) = 10C10*0.62^10*0.38^0 = 0.0084

c) P(X <= 4) = P(0) + P(1) + P(2) + P(3) + P(4) = 0.1348