Question

How can i find the probability of pulling a coin out of a jar? (details inside)

in a jar we have n coins, m of which are regular and the others give the result of H by a probability of `\frac{2}{3}`.
we pull a coin randomly and toss it 3 times - what are the chances that the coin we pulled is regular(m) if we know that we got at least 2H in 3 tossings?

please help me with that, i'm stuck and i don't know how to approach it

Answer 1

`(27m)/(40n-13m)`

Explanation:

For a regular coin, the probability of getting at least 2 heads out of 3 is the probability of getting 2 heads (`""^3C_2(1/2)^2(1/2) = 3/8`) plus the probability of getting 3 heads `(1/2)^3=1/8`, making a total of `1/2`.

For the biased coin, the probability of getting at least 2 heads is `""^3C_2(2/3)^2(1/3)+(2/3)^3 = 4/9+8/27=20/27`

Now, the probability that a coin drawn at random is regular is `m/n` , and that it is biased is `(1-m/n)`

We could use Bayes theorem to solve this problem in one step at this point. Let us instead try to gain some understanding by imagining a situation when this game is repeated a very large number `N` of times. We are assuming that `N` is so large that we can ignore the difference between relative frequency and probability, Out of the `N` games

• the number of times we draw a regular coin is `m/nN`
• out of these, the number of times we get at least 2 heads is `1/2m/nN`
• the number of times we draw a biased coin is `(1-m/n)N`
• out of these, the number of times we get at least two heads out of three is `20/27(1-m/n)N`

Thus, the total number of games in which there are at least 2 heads is

`1/2 m/n N + 20/27(1-m/n)N = (20/27-13/54 m/n)N`

Out of these the number of times the coin was regular is

`1/2 m/n N`

Thus, given that we did observe at least two heads, the probability that the coin is regular is

`( 1/2 m/n N)/((20/27-13/54 m/n)N)=(27m)/(40n-13m)`