Answer to Question #323799 in Statistics and Probability for Bless

Question

A multiple-choice examination has 15 questions, each with five possible answers, only one of which is correct. Suppose that one of the students who takes the examination answers each of the questions with an independent random guess. What is the probability that he answers at least ten questions correctly?

Answer 1

The probability that the student guesses the right answer $p = \frac{1}{5}=0.2,$ that he doesn't guess $q = \frac{4}{5}=0.8.$

Obviously the number of right answers X may be any value of 0, 1, 2, ... 14, 15.

We have a Bernoulli trial - exactly two possible outcomes, "success" (the student guesses the right answer) and "failure" (he doesn't guess) and the probability of success is the same every time the experiment is conducted (the student answers a question).

The probability of each result

$P(X=k)=\begin{pmatrix}n\\k\end{pmatrix}\cdot p^k\cdot q^{n-k}=\\ =\begin{pmatrix}15\\k\end{pmatrix}\cdot 0.2^k\cdot 0.8^{n-k}=\\ =\cfrac{15!}{k!\cdot(15-k)!}\cdot 0.2^k\cdot 0.8^{n-k}.$

The sought probability that he answers at least ten questions correctly:

$P(X\geq 10)=P(X=10)+P(X=11)+\\ +P(X=12)+P(X=13)+\\ +P(X=14)+P(X=15)=\\ =\cfrac{15!}{10!\cdot5!}\cdot 0.2^{10}\cdot 0.8^5+\cfrac{15!}{11!\cdot4!}\cdot 0.2^{11}\cdot 0.8^4+\\ +\cfrac{15!}{12!\cdot3!}\cdot 0.2^{12}\cdot 0.8^3+\cfrac{15!}{13!\cdot2!}\cdot 0.2^{13}\cdot 0.8^2+\\ +\cfrac{15!}{14!\cdot1!}\cdot 0.2^{14}\cdot 0.8^1+\cfrac{15!}{15!\cdot0!}\cdot 0.2^{15}\cdot 0.8^0=\\ =0.000113.$