Question

A certain medical test is known to detect 71% 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 = 0.71

Binomial probability distribution is

P(X) = ⁿC_x * p^x * ( 1 - p)^n-x

a)

P(X = 10) = ¹⁰C₁₀ * 0.71¹⁰ * ( 1 - 0.71)⁰

= 0.0326

b)

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

= ¹⁰C₈ * 0.71⁸ * ( 1 - 0.71)² + ¹⁰C₉ * 0.71⁹ * ( 1 - 0.71)¹ + ¹⁰C₁₀ * 0.71¹⁰ * ( 1 - 0.71)⁰

= 0.4099

c)

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

= ¹⁰C₀ * 0.71⁰ * ( 1 - 0.71)¹⁰ +¹⁰C₁ * 0.71¹ * ( 1 - 0.71)⁹ +¹⁰C₂ * 0.71² * ( 1 - 0.71)⁸ +¹⁰C₃ * 0.71³ * ( 1 - 0.71)⁷ +

¹⁰C₄ * 0.71⁴ * ( 1 - 0.71)⁶

= 0.0404