Question

A certain medical test is known to detect 59% 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:

Answer 1

`P(x = 3) = 0.048`

Explanation:

Given

`n = 10`

`p=59\% = 0.59`

Required

`P(x = 3)` --- probability that 3 are afflicted

This question illustrates binomial probability and it is calcuated using:

`P(x) = ^nC_x * p^x * (1 - p)^(n-x)`

So, we have:

`P(x = 3) = ^(10)C_3 * 0.59^3 * (1 - 0.59)^(10-3)`

`P(x = 3) = ^(10)C_3 * 0.59^3 * 0.41^7`

`P(x = 3) = 120 * 0.59^3 * 0.41^7`

`P(x = 3) = 0.048`