A multiple choice test consists of three problems. For each problem, there are five choices , one of which is correct .One student comes totally unprepared and decides to answer by sheer guessing . What is the probability that he will answer at least one problem correctly??
$\endgroup$ 12 Answers
$\begingroup$Denote $X$ = the number of correct answers.
$P(X\geq 1) = 1 - P(X=0)= 1 - \left(\dfrac{4}{5}\right)^3 = .488$
$\endgroup$ 3 $\begingroup$This has been answered but I got a completely different answer then DeepSea.
I used Number of Combinations, written en $nCr$ or $nCr(n, r)$.
$n$ = Number of questions.
$r$ = Chances of getting more than $X$ correct (1, 2, 3, 4 or 5 correct).
$a$ = Number of choices per question.
$\textbf{Let's fill the blanks!}$
$n = 5$
$r = {1, 2, 3, 4, 5}$
$a = 4$
$P(X \geq 0) = 1 - P(X < 1)$ $\textbf{..can also be written as..}$ $1 - P(X = 0)$
The probability of getting 1 or more correct out of 5 by chance with 4 choices per item
The formula is $nCr$ * $(chances-of-getting-correct)^r$ * $(chances-of-getting-wrong)^{n-r}$
$\hspace{17.1mm} =1 - nCr(5, 0) * (1/a)^0 * 3/a)^{5-0}$
$\hspace{17.1mm} =1 - nCr(5, 0) * (1/4)^0 * (3/4)^{5-0}$
$\hspace{17.1mm} =1 - 1 * 1 * 0.237305$
$\hspace{17.1mm} = 0.762695$
$\hspace{17.1mm} = \textbf{76.27%}$
But what about getting 2 or more correct? Simple!
1 - chances of getting 0 correct + chances of getting 1 correct.
$P(X \geq 1) = 1 - P(X < 2)$
$\hspace{17.1mm} =1 - (nCr(5, 0) * (1/4)^0 * (3/4)^{5-0}$
$\hspace{18.5mm} + nCr(5, 1) * (1/4)^1 * (3/4)^{5-1})$
$\hspace{17.1mm} = 1 - 0.632813$
$\hspace{17.1mm} = 0.367188$
$\hspace{17.1mm} = \textbf{36.72%}$
$\endgroup$
