Question

One of two coin is selected at random and tossed three times. the first coin comes up head with probability p1 = 1/3 and the second coin with probability p2 = 2/3
a. what is the probability that the number of heads is k = 0,1,2,3?
b. find the probability that coin 1 was tossed given that k heads were observed for k = 0,1,2,3.
c. In part b, which coin is more probable when k = 0,1,2,3 heads have been observed?

Answer 1

this should be easy for einstein :)

Answer 2

Einstein wants to test you

Answer 3

anyway, you need to use binomial theorem for each coin to see probability of getting 0,1,2,3 heads
\[P(X=k) = \left(\begin{matrix}n \\ k\end{matrix}\right) p^{k} (1-p)^{n-k}\]
n=3 , p = p1,p2 depending on which coin you pick
there is 50/50 chance you pick either coin
here is answer for k=0
\[P(k=0) = \frac{1}{2}\left(\begin{matrix}3 \\ 0\end{matrix}\right)(\frac{1}{3})^{0} (\frac{2}{3})^{3} +\frac{1}{2}\left(\begin{matrix}3 \\ 0\end{matrix}\right)(\frac{2}{3})^{0} (\frac{1}{3})^{3} \]
\[= \frac{1}{2}*\frac{8}{27} + \frac{1}{2}*\frac{1}{27}\]
\[= \frac{1}{6}\]

Answer 4

there is not 50/50 chance i pick either coin
first coin comes up head with probability p1 = 1/3 and the second coin with probability p2 = 2/3

Answer 5

Now for part b) you have to use conditional probability
\[P(A|B) = \frac{P(A and B)}{P(B)}\]
where A is coin 1 is picked , B is k heads are observed

Answer 6

it says "1 of 2 coins is selected at random" thats what i am referring to

Answer 7

the 2 coins are weighted different so that heads is either more or less likely ..... but that probability of picking either coin is 50%

Answer 8

here is answer to part b) for k=0
P(B) = P(k=0) = 1/6 as found in part a)

P(A and B) = probability that k=0 when using coin 1 (p=1/3)

\[P(A and B) = \frac{1}{2}\left(\begin{matrix}3 \\ 0\end{matrix}\right) (\frac{1}{3})^{0} (\frac{2}{3})^{3} = \frac{4}{27}\]
now you can find conditional probability
\[P(A|B) = \frac{4/27}{1/6} = \frac{4}{27}*\frac{6}{1} = \frac{8}{9}\]

Answer 9

now repeat for k=1,2,3

part c) follows from part b)
since for k=0 probability coin 1 is used is 8/9 , coin 1 is more probable