# Thread: Bucket/ Coin probability question

1. ## Bucket/ Coin probability question

There is a bucket containing 1000 coins.
1 of those coins has heads on both sides.
The other 999 are normal coins with heads and tails.

I pick one coin from random out of the bucket.

What is the probability that the coin I picked was the coin with heads on both sides?

I know the answer works out to roughly a half but I don't know how to work it out... any help?

Thanks

2. Originally Posted by emydawn
There is a bucket containing 1000 coins.
1 of those coins has heads on both sides.
The other 999 are normal coins with heads and tails.

I pick one coin from random out of the bucket.

What is the probability that the coin I picked was the coin with heads on both sides?

I know the answer works out to roughly a half but I don't know how to work it out... any help?

Thanks

Couldnt find the delete button but I typically worked it out just as I posted it, so dont worry about replying, but if youre interested, here's how it's done!

right so ive picked the coin, and its flipped heads 10 times in a row. therefore there are two situations that have happened.

1) ive picked a regular coin which has just happened to flip heads 10 times in a row

OR

2) ive picked the funny coin which will always flip heads anyway

so one of those situations has definitely happened. therefore u find those probabilities which work out to the following

1) ive picked a regular coin which has just happened to flip heads 10 times in a row
=999/1000 * 1/1024 = 0.000976

2) ive picked the funny coin which will always flip heads anyway
=0.001

sum the probabilities

which gives 0.001976

then i want the probaability that i DID pick the funny coin, which was 0.001

therefore the answer is 0.001 / 0.001976

which gives 0.506179

ta da

3. Exactly. You beat me to the punch. Glad you figured it out.

This is Bayes Theorem.

$P(A|B)=\frac{P(A)P(B|A)}{P(A)P(B|A)+P(A')P(B|A')}$

If we let A=the probability of getting the fake coin and B = the probability of getting 10 heads in a row, then:

$\frac{(\frac{1}{1000})(1)}{(\frac{1}{1000})(1)+(\f rac{999}{1000})(\frac{1}{1024})}=\frac{1024}{2023} \approx{0.5062}$