Question

A certain medical test is known to detect 68% 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

Answer:

Given,

p = 0.68

sample n = 10

Here X follows Binomial distribution

P(X) = nCr*p^r*q^(n-r)

nCr = n!/(n-r)!*r!

a)

P(X = 10) = 10C10*0.68^10*(1-0.68)^(10-10)

= 0.0211

b)

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

= 10C8*0.68^8*(1-0.68)^2 + 10C9*0.68^9*(1-0.68) + 10C10*0.68^10*(1-0.68)^0

= 0.2107 + 0.0995 + 0.0211

= 0.3313

c)

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

= 10C0*0.68^0*(1-0.68)^10 + 10C1*0.68^1*(1-0.68)^9 + 10C2*0.68^2*(1-0.68)^8 + 10C3*0.68^3*(1-0.68)^7 + 10C4*0.68^4*(1-0.68)^6

= 0.00001 + 0.0002 + 0.0023 + 0.0130 + 0.0482

= 0.0637