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). May be this is so easy but i m not getting how to do that. My question is.. In a party, 15 people were asked to select a number from 1 to 10, randomly. What is the probability 0f 3 out of 15 are choosing number 7. If we want to solve it using binomial then we have all the information required like n=15, p= 1/10, q= 9/10. Using this we can find P(3). What if I don't want to use binomial distribution. How can I solve this question then?

The way which I was trying is ...

Success,Failure,Number of cases


0, 15,1


1,14,15


2, 13, 105


3, 12, 455


4, 11, 1365


5, 10, 3003


6, 9, 5005


7, 8, 6435


8, 7, 6435


9, 6, 5005


10, 5, 3003


11, 4, 1365


12, 3, 455


13, 2, 105


14, 1, 15


15, 0, 1


    total number of cases= 32768                

According to me p(3)= 455*(1/ 32768)

But this answer is totally different from the answer, using binomial formula. I don't know where I am wrong. Plz help me :(

Thank you.

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1 Answer 1

up vote 1 down vote accepted

The count of $455$ is correct. However, "success" and "failure" are not equally likely. Dividing $455$ by $2^{15}$ essentially assumes that we are tossing a fair coin.

For example, the calculation assumes that $3$ successes and $12$ failures is just as likely as $12$ successes and $3$ failures. But it is clear that $12$ people choosing $7$ is extremely unlikely, while $3$ people choosing $7$ is not rare.

Added Record the choices of the $15$ people as a string of length $15$, made up of the "letters" $1$ to $10$. There are $10^{15}$ such strings, and they are all equally likely. We count the number of strings that have exactly $3$ $7$'s. The location of the $7$'s can be chosen in $\binom{15}{3}$ ways. Once we have done this, we have to fill the remaining $12$ slots with non-$7$'s. This can be done in $9^{12}$ ways. So our probability is $$\frac{\binom{15}{3}9^{12}}{10^{15}}.$$ Alternately and almost equivalently, the number of successes has Binomial Distribution. The probability of exactly $3$ successes is $\binom{15}{3}\left(\frac{1}{10}\right)^3\left(\frac{9}{10}\right)^{12}$.

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@Nicolas . Oh yes I didn't think in that way. They are not at all equally likely. But the same question again. what do I do now? –  Sh_sharma Apr 18 '13 at 7:35
    
@Sh_sharma: I thought the post indicated that you knew: we use the binomial distribution. The probability that the number of successes is $3$ is $\binom{15}{3}(1/10)^3(9/10)^{12}$. One can give a more combinatorial analysis, I think I will add it. –  André Nicolas Apr 18 '13 at 7:39
    
@Nicolas .. I know Binomial distribution but what I am asking is I don't want to use Binomial distribution. Is there any other way? Thanks for your replies :) –  Sh_sharma Apr 18 '13 at 8:15
    
Nicolas . Oh i am sorry I didn't see what u have added . Now its clear to me.. Thank you so much for your help :) –  Sh_sharma Apr 18 '13 at 8:22
    
One can avoid explicit mention of binomial coefficients by using permutations instead, or by using generating functions. In my added material, I did not use the Binomial distribution, I used counting. But all these ideas are close to each other. Your proposal that led to the wrong answer used a whole lot of binomial coefficients, in particular $\binom{15}{3}=455$. –  André Nicolas Apr 18 '13 at 8:28

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