Guy_that_Minors_in_Maths
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- Joined
- Nov 5, 2017
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- 4
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.