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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?

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    $\begingroup$ Poor Robert :( hopefully, next time he will sleep better. $\endgroup$ Dec 6 '21 at 4:07
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Good job. Your working seems fine.

You might like to state the probability distribution of $X$ explicitly.

$X\sim Bin(20, 0.2)$.

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