Question

A certain medical test is known to detect 71% of the people who are afflicted with...

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?

Homework Answers

Answer #1

X~ Bin ( n , p )

Where n = 10 , p = 0.71

Binomial probability distribution is

P(X) = nCx * px * ( 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

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