Coin toss probability: "There are 3 coins in a bag, 2 fair and 1 biased...."

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Coin toss probability: "There are 3 coins in a bag, 2 fair and 1 biased...."

Hi, I've been asked the following question:

"There are 3 coins in a bag. 2 of them are fair and 1 is biased. The fair ones, of course, have equal chances for coming up heads or tails. The biased coin is rigged in such a way that it has a 3/4 chance of coming up heads and 1/4 chance of coming up tails. A spectator randomly chooses a coin and flips it 3 times. if we know that he got the sequence "H, T, H", what are the chances that he chose a fair coin?"

Now I quickly calculated that this equals to (2/3*1/2*1/2*1/2)/[(2/3*1/2*1/2*1/2)+(1/3*3/4*1/4*3/4)] and that equals to 16/25.
My question is, if we only knew that after 3 flips, he ended up with 2 heads and 1 tails, yet if we didn't know the spesific sequence of them appearing, would the answer still be 16/25? Thanks.
 
Hi, I've been asked the following question:

"There are 3 coins in a bag. 2 of them are fair and 1 is biased. The fair ones, of course, have equal chances for coming up heads or tails. The biased coin is rigged in such a way that it has a 3/4 chance of coming up heads and 1/4 chance of coming up tails. A spectator randomly chooses a coin and flips it 3 times. if we know that he got the sequence "H, T, H", what are the chances that he chose a fair coin?"

Now I quickly calculated that this equals to (2/3*1/2*1/2*1/2)/[(2/3*1/2*1/2*1/2)+(1/3*3/4*1/4*3/4)] and that equals to 16/25.
My question is, if we only knew that after 3 flips, he ended up with 2 heads and 1 tails, yet if we didn't know the spesific sequence of them appearing, would the answer still be 16/25? Thanks.
The result would be the same, as you would expect.

There are three possible orderings for the outcome. If the coin is good, you will have a probability of \(\displaystyle 3/8\) instead of \(\displaystyle 1/8\); if the coin is bad, you will have a probability of \(\displaystyle 27/64\) instead of \(\displaystyle 9/64\). As you multiply both numerator and denominator by 3, the result is the same.
 
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