assume that random guesses are made for 8 multiple choice questions on an SAT test, so that there are n=9 trials, each with probability of success (correct) given by p=0.6. find the indicated probability for the number of correct answers. find the probability that the number x of correct answers is fewer than 4.

Accepted Solution

Answer:[tex]P(x<4)=0.0994[/tex]Step-by-step explanation:If we call x the number of correct questions obtained in the 9 attempts, then:x is a discrete random variable that can be modeled by a binomial probability distribution p, with n = 9 trials.So, the p of x successes has the following formula.[tex]P(x) =\frac{n!}{x!(n-x)!}*p^x(1-p)^{n-x}[/tex]Where:n = 9p = 0.6We are looking for P(x<4)By definition:[tex]P(x<4) = P(x\leq3) = P(0) + P(1) + P(2) +P(3)[/tex]Then:[tex]P(x\leq3)=\sum_{x=0}^{3} \frac{9!}{x!(9-x)!}*(0.6)^x(1-0.6)^{9-x}[/tex][tex]P(x\leq3)=0.0994[/tex]