Tricky coin probability

Question

Can someone please help, I am getting the wrong answer.

Consider four coins labelled as 1, 2, 3 and 4. Suppose that the probability of obtaining a ‘head’ in a single toss of the -h coin is $/4,\, = 1, 2, 3, 4.$ A coin is chosen uniformly at random and flipped.

Given that the flip resulted in a ‘head’, the conditional probability that the coin was labelled either 1 or 2 equals.

I tried doing this

P(coin 1 or 2 | tails) = P(coin 1 or 2 And Tails)/P(tails) \ =(1/4)(3/4)+(1/4)(2/4) / 1/4(3/4)+(1/4)(2/4)+(1/4)(1/4) =5/6

In improved notation: $$P(C_1 \cup C_2 | T) = P((C_1 \cup C_2)T)/P(T) \\ = \frac{(1/4)(3/4) + (1/4)(2/4)} {(1/4)(3/4) + (1/4)(2/4) + (1/4)(1/4)} = 5/6.$$

But this is the wrong answer as the answer is one of 1/10,2/10,3/10,4/10. How to do this?

Answer 1

Hint: The probability that a head is flipped is

$\sum_{i=1}^4 P(\text{"coin i is chosen"}) \cdot P(\text{"flip resulted in a ‘head’, "|"coin i is chosen"})$

$=\frac14\cdot \frac14+\frac14\cdot \frac24+\frac14\cdot \frac34+\frac14\cdot \frac44=\frac5{8}$

And the numerator is $(1/4)(\color{red}1/4) + (1/4)(2/4)$ since coin 1 and coin 2 are involved. You have used the probability of the third coin.

Therefore the answer of the question is $\frac{(1/4)(1/4) + (1/4)(2/4)}{\frac5{8}}=\frac{3/16}{5/8}=\frac{3}{10}$

Answer 2

First we find:$$P(H)=\sum_{i=1}^4P(H\mid C_i)P(C_i)=\frac14\frac14+\frac24\frac14+\frac34\frac14+\frac44\frac14=\frac{10}{16}=\frac58\tag1$$

Then:

$$P(C_1\cup C_2\mid H)P(H)=P((C_1\cup C_2)\cap H)=$$$$P(C_1\cap H)+P(C_2\cap H)=P(H\mid C_1)P(C_1)+P(H\mid C_2)P(C_2)=\frac14\frac14+\frac12\frac14\tag2$$

Combining $(1)$ and $(2)$ we find: $$P(C_1\cup C_2\mid H)=\frac85\left(\frac14\frac14+\frac12\frac14\right)=\frac3{10}$$