Conditional probability about biased coin

Question

Suppose there are two coins, with coin 1 landing heads when flipped with probability 0.3 and coin 2 with probability 0.5. Suppose also that we randomly select one of these coins and then continually flip it. Let $H_j$ denote the event that flip j,j≥1, lands heads. Also, let $C_i$ be the event that coin i was chosen, i=1,2.

I am having trouble to understand what is $P(H_2 | H_1)$

My idea is that $P(H_2|H_1)$ = $(P(H1|H2)*P(H2))/P(H1)$ It is easy to undertand $P(H1)$ and $P(H2)$ and it is easy to calculate them. But what is $P(H1|H2)$? I mean, if the coin lands head at the second flip, it is intuitive that the first flip is a tail and it should be zero.

Answer 1

First, you need to compute the relative probabilities that it was $~(C_1 ~: ~0.3 ~\text{Heads})~$ versus $~(C_2 ~: ~0.5 ~\text{Heads})~$ that produced the Heads on the first coin flip.

Based on conditional probability, the probability that it was coin $~C_1~$ is,

$$\frac{1/2 \times 0.3}{(1/2 \times 0.3) + (1/2 \times 0.5)} = \frac{3}{8}.$$

So, as a result of a Heads showing on the first coin flip, the $~(C_1:C_2)~$ probabilities have changed from $~(1/2:1/2)~$ to $~(3/8,5/8).$

Therefore, the probability of a Heads on the second coin flip is

$$(3/8 \times 0.3) + (5/8 \times 0.5) = \frac{34}{80}.$$

Answer 2

My idea is that $P(H_2\mid H_1) = (P(H_1\mid H_2)*P(H_2))/P(H_1)$

Well, yes, that is true. However...

It is easy to undertand $P(H_1)$ and $P(H_2)$ and it is easy to calculate them.

Indeed. It is a simple matter of using the Law of Total Probability.

But what is $P(H_1\mid H_2)$?

You should have found that $P(H_1)=P(H_2)$ . Therefore $P(H_1\mid H_2)=P(H_2\mid H_1)$ .

So, this approach is of no help. You shall have to do something slightly different.

I mean, if the coin lands head at the second flip, it is intuitive that the first flip is a tail and it should be zero.

No. There is no reason the first flip of a coin shall affect the second flip of the same coin. (Note: here 'flip' means 'toss high in the air and see how it lands' rather than 'gently turn over'.)

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Use the definition for Conditional Probability and the Law of Total Probability.

$$\begin{align}P(H_2\mid H_1) &= \dfrac{P(H_1, H_2, C_1)+P(H_1, H_2, C_2)}{P(H_1, C_1)+P(H_1, C_2)}\end{align}$$

Also note that the events of $H_1$ and $H_2$ shall be conditionally independent for a given coin.