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?
Thanks!
(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 $p>1/2$).