Question

You have a loaded coin that comes up heads with a 0.6 probability. Now you add a fair coin
(probability heads = 0.5). You now have two coins - one loaded and one fair. You select a coin at
random with a probability of 0.5, you flip it twice and it comes up heads twice.
What is the probability that you picked the loaded coin?

Answer 1

Hint: follow conditional probability theorem:

Answer 2

The probability that the following is the answer is 0.6 :\[\frac{\frac{1}{2}\left(\frac{6}{10}\right)^2}{\frac{1}{2}\left(\frac{6}{10}\right)^2+\frac{1}{2}\left(\frac{1}{2}\right)^2}=\frac{36}{61} \]

Answer 3

You have two probabilities:

In coin 1 The probability that two heads come up. = 1/4 = 25/100
In coin 2 The probability that two heads come up. = 9/25 = 36/100

so you can conclude that: 9/25 > 1/4

You pick either randomly 50-50
What you want to know that according to the evidence that 2 heads came up from your chosen coin what is the likeliness that this chosen coin is the 9/25 coin.

You can tell by simple logic that as 9/25 > 1/4, the loaded coin is more likely to be the one you chose as it obeys the probabilities more than the other coin.

The exact number I don't know how to get it, I speculate it must be some sort of sum or difference between these two numbers ( 9/25, 1/4)

maybe 9/25+1/4 = 61/100

from here sincerely I don't know, I have never heard of such problem on probabilities where you use posterior evidence to calculate previous decisions