Bayes Theorem on Urnes with Coinflipping

Question

I could need your help with this. I got a urn with 2 coins, C1 and C2. Both have Head and Tails on it.

• P(C1=head) = 0.2
• P(C1=tails) = 0.8
• P(C2=head) = 0.4
• P(C2=tails) = 0.6

This is given. Now I flip 3 times and get in total 2 head and 1 tails. After those results, what is the probability, that I flipped Coin C1?

All i can think of, is the probability to get those relusts, which should be 0.2x0.2x0.8 = 0.032. But that doesn't help me any further. I am supposed to take at most 5 minutes to solve this and i think i am supposed to use Bayes Theorem, if it works here. But i don't have any clue.

Answer 1

The probability to get those results for coin $C_1$ is not the $0.2\times 0.2\times 0.8$ but three time higher. The event that we obtained is that we get two heads and one tail in total, not necessarily in this order. There are $3$ ways to arrange two heads and one tail $HHT$, $HTH$, $THH$, so $$ \mathbb P(2H+1T \mid C_1) = \binom{3}{1}\times 0.2\times 0.2\times 0.8. $$ You can also look at binomial distribution.

Calculate the same way $ \mathbb P(2H+1T\mid C_2) $ and then use Bayes formula with $\mathbb P(C_1)=\mathbb P(C_2)=\frac12$.

Answer 2

1. The first step is to find the prob of the outcome for each coin. You got "in total 2H, 1T", so you might have got them in any order. There are 8 possible outcomes for 3 flips, and three of those give 2H, 1T, namely HHT, HTH, THH. So, calling the event "in total 2H,1T" $E$ we have $$p(E|C_1)=3\times0.2^2\times0.8=0.096$$ Similarly, $$p(E|C_2)=3\times0.4^2\times0.6=0.288$$

2. The second step is to apply Bayes. Your prior distribution is $p(C_1)=p(C_2)-0.5$, So Bayes gives $$p(C_1|E)=\frac{0.096}{0.096+0.288}=0,25$$

Answer 3

Bayes in short.

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Observe that: $$P(C_1\mid T=2)P(T=2)=P(T=2\mid C_1)P(C_1)\tag1$$ Find $P(T=2)$, $P(T=2\mid C_1)$ and $P(C_1)$ and substitute in $(1)$.

Then the equality enables you to find $P(C_1\mid T=2)$.