Question

A quiz consists of 6 multiple choice questions.each question has 4 possible answers.a student is unprepared and guess answers by random.he passes if he gets atleast 3 questions right.what is the probability that he will pass

Answer 1

Thus, simply compute the probabilities for 4,5,6,7 as well, and then add. Alternatively, you can compute the probabilities for 0,1,2 correct answers, and subtract the total from 1, giving the same result.

Answer 2

16.94% probability that he will pass

Explanation:

For each question there are only two possible outcomes. Either the answer it correctly, or he does not. The probability of answering 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) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

6 questions

So n = 6.

4 possible answers

So
`p = (1)/(4) = 0.25`

He passes if he gets atleast 3 questions right.what is the probability that he will pass

`P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)`

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 3) = C_(6,3).(0.25)^(3).(0.75)^(3) = 0.1318`

`P(X = 4) = C_(6,4).(0.25)^(4).(0.75)^(2) = 0.0330`

`P(X = 5) = C_(6,5).(0.25)^(5).(0.75)^(1) = 0.0044`

`P(X = 6) = C_(6,6).(0.25)^(6).(0.75)^(0) = 0.0002`

`P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.1318 + 0.0330 + 0.0044 + 0.0002 = 0.1694`

16.94% probability that he will pass