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...2 to 5 questions answered correctly, out of 20 of them? Each question has 5 choices. Probability of getting one right is 1/5. Probability of getting exactly 1 right is ${20 \choose 1} p^1 q^{19}$, with $p=P(\mathrm{right})$ and $q=P(\mathrm{wrong})$ (which I managed to understand and calculate). However how do I calculate for the problem above?

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Thanks for the "TeXation" mpiktas :)) – Queops Jan 27 '11 at 13:55
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your are welcome. Note that right clicking on formula will present you pop-up menu with a choice to see the source. – mpiktas Jan 27 '11 at 13:57
@mpiktas: Nice to know that it is possible to see the source. – Tomek Tarczynski Jan 27 '11 at 14:11
If you feel @mpiktas's answer helped you to solve your problem, then the best way to thank him is probably to upvote his response. – chl♦ Jan 27 '11 at 14:41

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Hint: sum up the probabilities. The probability that exactly $k$ answers are answered correctly is $${20 \choose k}\left(\frac{1}{5}\right)^k\left(\frac{4}{5}\right)^{20-k}.$$ In your case you have $k=2,3,4,5$.

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You confirm the answer is aprox. 0.735 ? Seems the problem solution might be wrong. – Queops Jan 27 '11 at 14:03
@Queops, R code pbinom(5,20,1/5)-pbinom(1,20,1/5) gives 0.7350325, so yes. – mpiktas Jan 27 '11 at 14:07
@Queops, maybe the problem is to calculate the probability that exactly questions numbered 2 to 5 are answered correctly? Then probability should be calculated differently. – mpiktas Jan 27 '11 at 14:08
No no, it's just the number, doesn't matter which ones. And here I was trying to figure out what was wrong. Thanks so much for the help. – Queops Jan 27 '11 at 14:10

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