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Coin $i$ has probability $P_i=\frac{i}{10}$ of heads, where $i=1,...,9$. A coin is randomly chosen (each of the 9 coins is equally likely to be chosen), and the randomly chosen coin is flipped 4 times, resulting in 3 heads and 1 tail. Find the conditional probability that coin 5 was the chosen coin. Hint: Bayes rule

My Attempt:

Let's define T = getting the given flip combination

Let's define 5 = getting coin 5

Following from Bayes' Formula: $P(5 \mid T)=\frac{P(T\mid 5)P(5)}{P(T \mid 5)P(5)+P(T \mid 5^\mathsf{c})P(5^\mathsf{c})}$

$P(5^\mathsf{c})=\frac{8}{9}$

P(coin 1 getting the testing combination) $=(\frac{1}{10})^3\times\frac{9}{10}$

This is where I'm confused. Is the below correct? Because for each coin, the probability of choosing that coin is $\frac{1}{9}$, and then we multiply by the probability of getting the given combination.

$P(T and 5^\mathsf{c})=\frac{1}{9}\times[(\frac{1}{10})^3\times\frac{9}{10}+(\frac{2}{10})^3\times\frac{8}{10}+(\frac{3}{10})^3\times\frac{7}{10}+(\frac{4}{10})^3\times\frac{6}{10}+(\frac{6}{10})^3\times\frac{4}{10}+(\frac{7}{10})^3\times\frac{3}{10}+(\frac{8}{10})^3\times\frac{2}{10}+(\frac{9}{10})^3\times\frac{1}{10}]=something$

I'm very confused. Thanks for any help!

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    At first glance, that actually all looks to be correct! – 2017-02-13
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    Ah thanks for the prompt response. But then would it be possible to explain the intuition behind P(T and 5^c)? Me doing this problem was like stumbling around in the dark. – 2017-02-13
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    First ... i did spot a small mistake: in your last formula the demoninator should be $P(T \land 5) + P(T \land 5^C) = 0.0625 + something$. That is, since $P(T|5^C) =something/8/9$ and $P(5^C) = 8/9$ the $8/9$ cancels out. – 2017-02-13
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    Also, technically, to compute the probability of getting 3 heads and 1 tail, you would have to take into account all 4 ways of getting that, i.e. The 1 tail can be one of the four throws. Thus, you would have to multiply each of the $P(T \land i)$ by 4 ... But in the end all that means is that in the last formula you get a 4 in both the numerator and the ddenominator, so they cancel out. – 2017-02-13
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    Oh wow, I didn't realize, thanks for the note! – 2017-02-13

1 Answers 1

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Yes, you did this right!

To explain $P(T \land 5^C)$:

$P(T \land 5^C) = P(T \land (1 \lor 2 \lor 3 \lor 4 \lor 6 \lor ... \lor 9)) = P((T \land 1) \lor (T \land 2) \lor ... \lor (T \land 9)) = P(T \land 1) + P(T \land 2) + ... + P(T \land 9)$

The first steps are all pure logic, and the last step is because $T \land i$ And $T \land j$ with $i \not = j$ are mutually exclusive events.

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    Thank you!! I was bumbling around without knowing why, this is very helpful! – 2017-02-13
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    @Wren You're welcome! :) – 2017-02-13