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A factory makes three types of biased coins with probability of getting head when tossed is $\frac{1}{2}$,$\frac{1}{3}$,$\frac{1}{4}$ respectively. Each coin is tossed before packing and the face that appears is marked.

(1) A coin is selected from a bag containing one coin of each type and the coin is tossed and head appears. Find the probability that the face that appeared is marked.

(2) If a coin is selected from the bag and the 'heads' side is marked, the probability that the tossed coin will show heads is?

In the first case, the probability of choosing any type of coin is $\frac{1}{3}$

The probability of 'head' being marked in each type of coin is, $\frac{1}{2}$,$\frac{1}{3}$,$\frac{1}{4}$

The probability of getting heads is $\frac{1}{2}$,$\frac{1}{3}$,$\frac{1}{4}$

There are three possible cases. The total probability is $$\frac{1}{3}\times\frac{1}{2}\times\frac{1}{2}+\frac{1}{3}\times\frac{1}{3}\times\frac{1}{3}+\frac{1}{3}\times\frac{1}{4}\times\frac{1}{4}$$

But the answer I got was wrong. The answer given is $\frac{61}{12\times 13}$.

(I can get the answer if I divide the answer I got, by the probability of getting a head irrespective of the mark.)

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  • $\begingroup$ It seems to be correct though. $\endgroup$ – SinTan1729 Feb 19 '16 at 10:19
  • $\begingroup$ You should include the given answer when you know it. $\endgroup$ – Em. Feb 19 '16 at 10:24
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Your last remark is indeed correct. The probability you calculated is $P(H \cap H_M)$, the probability that a coin from those 3 has its head marked and comes up $H$.

$P(H) = \frac{1}{3}(\frac{1}{2} + \frac{1}{3} + \frac{1}{4}) = \frac{13}{36}$, by conditioning again.

The asked for probability is (head is marked given that head came up!): $P(H_M | H) = {P(H \cap H_M) \over P(H)}$ and then we get the answer $\frac{61}{3 \cdot 12 \cdot 12} \cdot \frac{36}{13} = \frac{61}{12 \cdot 13}$ as required.

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