Question

A certain medical test is known to detect 51% 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

X ~ bin ( n , p)

Where n = 10 , p = 0.51

binomial probability distribution is

P(X) = ⁿC_x p^x ( 1 - p)^n-x

a)

P(X = 10) = ¹⁰C₁₀ * 0.51¹⁰ * ( 1 - 0.51)⁰

= 0.0012

b)

P(X >= 8) = P(X = 8) + P(X = 9) + P(X = 10)

= ¹⁰C₈ * 0.51⁸ * ( 1 - 0.51)² +¹⁰C₉ * 0.51⁹ * ( 1 - 0.51)¹ +¹⁰C₁₀ * 0.51¹⁰ * ( 1 - 0.51)⁰

= 0.0621

c)

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

= ¹⁰C₀ * 0.51⁰ * ( 1 - 0.51)¹⁰ +¹⁰C₁ * 0.51¹ * ( 1 - 0.51)⁹ +¹⁰C₂ * 0.51² * ( 1 - 0.51)⁸

+¹⁰C₃ * 0.51³ * ( 1 - 0.51)⁷ +¹⁰C₄ * 0.51⁴ * ( 1 - 0.51)⁶

= 0.3526