In a medical examination, the chances of error are 15%. Now, find the Bernoulli distribution if one patient is randomly selected out of 60 patients.?

P(X=x) = \$p^x q^(1-x)\$
where x =0,1
q=1-p

Number of error reports when 60 patients are examined

15% of 60 = 9

Number of patients getting the correct reports

60 - 9 = 51

Given probability

p = \$51/60 = 17/20\$

Then q = 1-p

q = \$1 -(17/20)\$

Substitute x=0

P(X=0) = \$(17/20)^0 (1 -(17/20))^(1-0)\$

P(X=0)

\$1 -(17/20) = 3/20\$ = 0.15

Substitute x=1

P(X=1) = \$(17/20)^1 (1 -(17/20))^(1-1)\$

P(X=1)

\$(17/20)\$

Probability of successful result in medical test

0.85

probability of failure or error

0.15

Answer

A