Question

A test consists of 10 true/false questions. To pass the test a student must answer at least 6 questions correctly. If a student guesses on each question, what is the probability that the student will pass the test?

Answer 1

Answer:

37.70% probability that the student will pass the test

Step-by-step explanation:

For each question, there are only two possible outcomes. Either the student guesses it correctly, or he does not. The probability of a student guessing a question correctly is independent of other questions. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

10 true/false questions.

10 questions, so $n = 10$

True/false questions, 2 options, one of which is correct. So $p = \frac{1}{2} = 0.5$

If a student guesses on each question, what is the probability that the student will pass the test?

$P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 6) = C_{10,6}.(0.5)^{6}.(0.5)^{4} = 0.2051$

$P(X = 7) = C_{10,7}.(0.5)^{7}.(0.5)^{3} = 0.1172$

$P(X = 8) = C_{10,8}.(0.5)^{8}.(0.5)^{2} = 0.0439$

$P(X = 9) = C_{10,9}.(0.5)^{9}.(0.5)^{1} = 0.0098$

$P(X = 10) = C_{10,10}.(0.5)^{10}.(0.5)^{0} = 0.0010$

$P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.2051 + 0.1172 + 0.0439 + 0.0098 + 0.0010 = 0.3770$

37.70% probability that the student will pass the test