Answer:
2.78% probability that he gets exactly 2 of the twenty questions wrong
Explanation:
For each question, there are only two possible outcomes. Either he knows the answer, or he does not. The probability of him knowing the answer for a question 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)](https://img.qammunity.org/2021/formulas/mathematics/college/mj488d1yx012m85w10rpw59rwq0s5qv1dq.png)
In which
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)!)](https://img.qammunity.org/2021/formulas/mathematics/college/qaowm9lzn4vyb0kbgc2ooqh7fbldb6dkwq.png)
And p is the probability of X happening.
On any question, the student knows the answer with probability .7.
This means that
![p = 0.7](https://img.qammunity.org/2021/formulas/mathematics/high-school/mgkca71qc5cdp502iri38ylfbb5mvcnnji.png)
What is the probability that he gets exactly 2 of the twenty questions wrong, if his performance on different questions is independent
2 of 20 wrong, 20-2 = 18 correctly. So this is P(X = 18).
![P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)](https://img.qammunity.org/2021/formulas/mathematics/college/mj488d1yx012m85w10rpw59rwq0s5qv1dq.png)
![P(X = 18) = C_(20,18).(0.7)^(18).(0.3)^(2) = 0.0278](https://img.qammunity.org/2021/formulas/mathematics/college/872tspsnqoo8o07z5fm83k333f0v5nj9sc.png)
2.78% probability that he gets exactly 2 of the twenty questions wrong