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Once more I have a problem with probability exercise.

Probability that an item is a standard one is equal to 0.8. Find the probability that, when choosing two items only one of them is of standard type.

An aswer given in the book is 0.18

Shouldn't it rather be 2*0.8*0.2=0.32?

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You are right based on what you have described. Since, $P$(one is standard)=$1-P$(two are standard)$-P$(none are standard) which comes out to be 0.32 as well. Are you sure you're stating the problem correctly here? –  jay-sun Feb 26 '13 at 17:35
    
If the probability truly is $0.18$ then $2p(1-p)=0.18$ and so $p=0.1$ or $p=0.9$. –  Fly by Night Feb 26 '13 at 17:40
    
I think it is just a print mistake, because this book is really good. But it is always better to see what other ppl think about it. And yes it is exactly stated this way, and translated almost word by word. :] Thanks guys. –  Misery Feb 26 '13 at 18:05
    
@Misery which book? –  jay-sun Feb 26 '13 at 18:27
    
@jay-sun: W.J. Gmuman Exercises on probability and statistics (Polish ed.) –  Misery Mar 6 '13 at 15:42

1 Answer 1

up vote 1 down vote accepted

Assuming there are a large enough sample such that each selection is independent (i.e. choosing a standard item does not affect the probability that the next selection is a standard item) you can model a problem like this using a binomial distribution.

The random variable $X$ (the number of standard type items selected) is distributed binomially with $n=2$ (two selections) and $p=0.8$ (probability of success, i.e. choosing an item of standard type with each selection is $0.8$). That is: $X\sim B(2,0.8)$.

Therefore, using the P.M.F for the binomial distribution, we have:

$$P(X=1)={2\choose 1}\times0.8\times0.2=0.32$$

Which is the answer that you have obtained, so the book is indeed mistaken.

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Thanks, but still, there is not mentioned that the sample is large. But now I know what the author had on his mind :] –  Misery Feb 26 '13 at 18:10

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