Question

A certain medical test is known to detect 75% 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 that p = 0.75
n = 10
formula
p(X = x) = ncr * p^r * q^(n-r)
=> All 10 have the disease
p(x = 10) = 10c10 * (0.75)^10 * (0.25)^(0)
= 0.0563

=> At least 8 have the disease
p(x ≥ 8) = p(x = 8) + p(x = 9) + px = 10)
= 10c8 * (0.75)^8 * (0.25)^2 + 10c9 * (0.75)^9 * (0.25)^1 + 10c10 * (0.75)^10 * (0.25)^0
= 0.2816+0.1877+0.0563
0.5256

=> At most 4 have the disease
p(x ≤ 4) = p(x = 0) + p(x = 1)+p(x = 2) + p(x = 3) + p(x = 4)
= 10c0 *(0.75)^0 * (0.25)^10 + 10c1 *(0.75)^1 * (0.25)^9 + 10c2 *(0.75)^2 * (0.25)^8 + 10c3 *(0.75)^3 * (0.25)^7 + 10c4 *(0.75)^4 * (0.25)^6
= 0.0000+0.0000+0.0004+0.0031+0.0162
= 0.0197