Tossing two coins one fair and one double tailed.

Question

If you toss a coin randomly from two coins

1. Fair coin
2. Double tailed

If you got tails then what is the probability of heads being on the other side of same coin?

My approach:

Probably of picking fair or biased coin is 50/50.

Case I: fair coin is picked. Getting heads on other side is 1. Since, we already got tails

Case II: biased coin is picked Probability of getting heads is 0 since there is no heads.

Probably of the resulting probably to be 1 is 50% and vice versa.

Actual source answer. But when I looked it said 1/3.

Can anyone explain why my approach is wrong?

Answer 1

I remember that even I used to struggle with such questions.
Here, the key idea is that you are already given that a tail is observed. This does not leave a $0.5$ probability to each coin. This is because the second coin has $2$ tails. So, it has a probability of $2/3$, while the other has $1/3$. So, calculating, we get: $$P = (2/3)0+(1/3)1 = \boxed{1/3}$$

This can be explained by considering a $4$ sided dice (with numbers $1-4$) and a $100$ sided die with numbers $1-100$. If I tell you that I randomly picked a dice and rolled it to get $1$, will you say that the probability that I picked the four sided dice is $0.5$?

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We can directly calculate using the conditional probability formula. $$A = \{\text{there is a heads on the other side of the coin}\}$$ $$B = \{\text{tails}\}$$ Notice that $A \subset B$ so $A \cap B = A$. $$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A)}{P(B)} = \frac{1/4}{3/4} = \boxed{1/3}$$

Answer 2

There are four faces in total, each of which could show up with equal probability.

But only three of these faces may be the tail you do see.

And just one of these tails is on a coin with a head on the other side.