Robert didn’t sleep last night and now he is sitting for a multiple choice examination with ...

Robert didn’t sleep last night and now he is sitting for a multiple choice examination with 20 questions. He fills in the answers on the bubble sheet randomly without reading the question and each question has 5 possible answers.

A.) What is the probability that Robert gets exactly 10 questions right?

• $$P(X=10) = \binom{20}{10}\left( \frac15\right)^{10}\cdot \left(\frac45\right)^{10} = 0.002$$

B.) What is the probability that Robert gets two or more questions right?

• $$P(x \ge 2) = 1 - P(x<2) = 1 - [P(x=0) + P(x=1)] = 0.9308$$

c.) What is the expected value and variance of the number of questions Roberts gets right?

• $$E(x) = 20(1/5) = 4$$

$$Var(x) = 20(1/5)(4/5) = 16/5$$

Does my thought process seem correct or does my work or my answers seem off somewhere?

• Poor Robert :( hopefully, next time he will sleep better. Dec 6 '21 at 4:07

You might like to state the probability distribution of $$X$$ explicitly.
$$X\sim Bin(20, 0.2)$$.