I had an exam and this was one of the questions, I was wondering if I was fully correct or not, because I kind of took a while on it and was unsure. There are three parts to it.
The question is (from what I can recall):

A coin experiment is conducted. There are three coins in a bag, copper, silver, gold. There is $\frac{1}{3}$ chance of pulling out any of the coins. Once we pull out a coin, we will flip that coin $10$ times and record how many heads appears. Let $N$ be the random variable such that it counts the number of heads. The probability of flipping heads for each coin is $0.7,0.5,0.3$ respectively for each coin.
i) What is the probability of getting no heads (i.e. $N=0$)?
This is what I did:
$\mathbb{P}(N=0) = \frac{1}{3} (0.7^{10} + 0.5^{10} + 0.3^{10})$

ii) What is the probability of getting $n$ heads (i.e. $N=n$)?
This is what I did:
$\mathbb{P}(N=n) = \frac{1}{3} \begin{pmatrix}10 \\ n \end{pmatrix}(0.7^n 0.3^{10-n} + 0.5^{n}0.5^{10=n} + 0.3^{n}0.7^{10-n})$.

(iii) Is the distribution of $N$ binomial?
I said:
No. Because even though the sum of binomials is binomial, we have a constant dividing this binomial. A binomial divided by a constant is not necessarily binomial.

Are these all corect and if so/not, how would I do this/do this better?


  • $\begingroup$ When they say "the sum of binomials is binomial", that's not what they mean. They mean if you have a single coin (fair or not), and flip it 10 times, then the number of heads is binomially distributed. If you after that flip it 15 times, then the number of heads in this second experiment is also binomially distributed. If you add the heads from the first and the second experiment (i.e. take the sum of the two binomial variables), then the result is binomially distributed as well. Intuitively quite an obvious result, wouldn't you say? $\endgroup$ – Arthur Jun 19 '17 at 8:42
  • $\begingroup$ Sorry i might be misunderstanding you, isn't the part ii our probability distribution function for $N$, so it won't be binomial since there isn't any possible binomial r.v. with that kind of function? And yep, the adding of binomials seems intuitive to me because it's the addition of bernoulli trials (if they have the same parameter) I think? $\endgroup$ – Natash1 Jun 19 '17 at 9:12

(i) is correct. Also the answer of (ii) is correct, but the reason is that $$ \frac{a^i(1-a)^{10-i}+b^i(1-b)^{10-i}+c^i(1-c)^{10-i}}{3} \neq p^i(1-p)^{10-i} $$ if you fix $0<a<b<c<1$ and $p \in (0,1)$ for all $i=0,1,\ldots,10$. Indeed, if $i=10$ you have $$ p=\left(\frac{a^{10}+b^{10}+c^{10}}{3}\right)^{1/10}. $$ Note that by Jensen inequality $p>\frac{1}{3}(a+b+c)=1/2$. Moreover if $i=0$ you have $$ p=1-\left(\frac{(1-a)^{10}+(1-b)^{10}+(1-c)^{10}}{3}\right)^{1/10}. $$ In your case $\{a,b,c\}=\{1-a,1-b,1-c\}$ hence the two expressions cannot be equal unless $p=1/2$ (but we have shown $p>1/2$).

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  • $\begingroup$ Thank you! I don't get where you got your first "$\neq$" though, it looks like the result in part $ii$ except, with no binomial coefficient? I understand the rest though (when accepting your first line) $\endgroup$ – Natash1 Jun 19 '17 at 9:10
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    $\begingroup$ The binonial coefficient is constant in both distribution, if they were the same, hence they cancel out. The remaining part is just a proof of that "\neq": suppose for the sake of contradiction that they are equal for all i, then... $\endgroup$ – Paolo Leonetti Jun 19 '17 at 12:17

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