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I am trying to understand what is the implied distribution of the following problem: A student asks for help from the professor 2 times per test on average. the student took 5 different test

a) what is the probability that at only 2 tests the student didn't ask from help at all

b) the probability that in 3 tests at the most the student asked for help only 3 times

I know that if Poisson distribution is not specified then it cant be assumed. I am at a loss on this one. Help will be appreciated.

Edit: How would i solve it assuming Poisson distribution?

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2 Answers 2

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You're right, there's not enough information to solve the problem; one would have to add an assumption about the distribution (as you say, most likely a Poisson distribution) to solve it.

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  • $\begingroup$ How would i solve it assuming Poisson distribution? $\endgroup$
    – SteelSoul
    May 9, 2013 at 0:24
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Assuming Poisson distribution,

a) $\displaystyle P(x)=\frac{\mu^xe^{-\mu}}{x!}$

The probability that a student doesn't ask for help is when $\mu=2$ and $x=0$

$\displaystyle P(0)=\frac{(2\times 2)^0e^{-(2\times2)}}{0!}=e^{-4}\approx 0.0183$

b) The probability that a student asks for help thrice in at most 3 tests out of 5 is given by the binomial distribution,

$\displaystyle P(X\leq3)=\sum_{X=0}^3 \frac{(2\times 3)^Xe^{-(2\times 3)}}{X!}\approx0.1512$

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  • $\begingroup$ Very simple. The summing of b eluded me. Thank you for the explanation. $\endgroup$
    – SteelSoul
    May 9, 2013 at 2:25

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