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

1 Answer
Apr 21, 2018

(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)