Question

A student takes an exam containing 18 multiple choice questions. The probability of choosing a correct answer by knowledgeable guessing is 0.3. If the student makes knowledgeable guesses, what is the probability that he will get between 8 and 12 (both inclusive) questions right? Round your answer to four decimal places.

Answer 1

When you have to repeatedly take the same test, with constant probability of succeeding/failing, you have to use Bernoulli's distribution. It states that, if you take $n$ tests with "succeeding" probability $p$, and you want to "succeed" k of those n times, the probability is

$\displaystyle P(n,k,p) = \binom{n}{k}p^k(1-p)^{n-k}$

In your case, you have n=18 (the number of tests), and p=0.3 (the probability of succeeding). We want to succeed between 8 and 12 times, which means choosing k=8,9,10,11, or 12. For example, the probability of succeeding 8 times is

$\displaystyle P(18,8,0.3) = \binom{18}{8}(0.3)^8(0.7)^{10}$

you can plug the different values of k to get the probabilities of succeeding 9, 10, 11 and 12 times, and your final answer will be

$P = P(18,8,0.3) + P(18,9,0.3) + P(18,10,0.3) + P(18,11,0.3) + P(18,12,0.3)$