Question

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.

Answer 1

Answer:

$P(x$

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.

$P(x) =\frac{n!}{x!(n-x)!}*p^x(1-p)^{n-x}$

Where:

n = 9

p = 0.6

We are looking for P(x<4)

By definition:

$P(x$

Then:

$P(x\leq3)=\sum_{x=0}^{3} \frac{9!}{x!(9-x)!}*(0.6)^x(1-0.6)^{9-x}$

$P(x\leq3)=0.0994$