Question

5. We have 3 coins: 1 normal coin, 1 trick coin with 2 heads and 1 trick coin with 2 tails. A coin is drawn at random and put on a table and you see that it is

heads. What is the probability you have the real coin? (4 points)

Answer 1

1/3

Step-by-step explanation:

Imagine you repeat the experiment 100 times.

It is expected that half of the times (50) you take the fair coin and half of the times (50) the trick coin.

When you take the fair coin, half of the times you get heads, that is 25 times from 100 you “get the fair coin AND you get heads with the fair coin”.

When you take the trick coin, all the times you get heads, that is 50 times from the original 100 you “get the trick coin AND you get heads with the trick coin”.

In total, it is expected you get heads 75 times. And from those, 50 times will be with the trick coin and 25 times with the fair coin.

Now, if you get one of those at random, what’s the likelihood it was the fair coin?

Well, you have 25 cases it was the first coin and 50 cases it wasn’t, so it is 25 from a total of 75 and that is 25/75 = 1/3. That is, about 33.33%

Also, with probability theory:

Bayes’ Theorem:

P(B|A) = P(A ∩ B) / P(A)

P(A|B) = P(A ∩ B) / P(B)

So: with just the definitions I prove Bayes:

P(B|A) = P(A|B) * P(B) / P(A)

In our case:

P(Fair| Heads) = P(Heads|Fair) * P(Fair) / P(Heads)

p = P(Fair | Heads) = 0.5 * 0.5 / P(Heads)

What is P(Heads) ??

We can apply this:

Heads = (Heads ∩ Fair) U (Heads ∩ Trick)

And those in the union are disjoint (mutually exclusive), so:

P(Heads) = P(Heads ∩ Fair) + P(Heads ∩ Trick) =

= P(Heads|Fair) * P(Fair) + P(Heads|Trick) * P(Trick) =

= 0.5 * 0.5 + 1.0 * 0.5 = 1/4 + 1/2 = 3/4

Then:

p = P(Fair | Heads) = 0.5 * 0.5 / P(Heads) = 1/4 / (3/4) = 1/3