...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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1$\begingroup$ your are welcome. Note that right clicking on formula will present you pop-up menu with a choice to see the source. $\endgroup$– mpiktasJan 27, 2011 at 13:57
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$\begingroup$ @mpiktas: Nice to know that it is possible to see the source. $\endgroup$– Tomek TarczynskiJan 27, 2011 at 14:11
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$\begingroup$ 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. $\endgroup$– chlJan 27, 2011 at 14:41
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1 Answer
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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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$\begingroup$ You confirm the answer is aprox. 0.735 ? Seems the problem solution might be wrong. $\endgroup$– QueopsJan 27, 2011 at 14:03
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$\begingroup$ @Queops, R code
pbinom(5,20,1/5)-pbinom(1,20,1/5)gives 0.7350325, so yes. $\endgroup$– mpiktasJan 27, 2011 at 14:07 -
$\begingroup$ @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. $\endgroup$– mpiktasJan 27, 2011 at 14:08
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$\begingroup$ 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. $\endgroup$– QueopsJan 27, 2011 at 14:10