Find the probability that the number x of correct answers is fewer than 4.

Question

Assume that random guesses are made for

nine

multiple choice questions on an SAT​ test, so that there are

n=9

​trials, each with probability of success​ (correct) given by

p=0.65.

Find the indicated probability for the number of correct answers.

Answer 1

P(<4 right) = P(0,1,2, or 3)

P(0) = 9C0(.35)^9 = .0000788

P(1) = 9C1(.65)(.35)^8 = 9(.65)(.000225)= .001317

P(2) = 9C2(.65)^2(.35)^7= .009786

P(3) = 9C3(.65)^3(.35)^6= .0424

sum them = 0.05 = about 5% chance of getting less than 4 right

p(9) = 9C9(.65)^9 = .0207

P(8) = 9C8(.35)(.65)^8 = 9(.35)(.65^8) = .10037

P(7) = 9C7)(.35^2)(.65^7) = .21619

P(6) = 9C6(.35^3)(.65^6) = .2716

P(5) = 9C5(.35^4)(.65^5) = .21939

P(4) = 9C4(.35^5)(.65^4) = .11813

sum = about .95 = 95% chance of 4 or more right

probability of less than 4 = 100%-95% = about 5%