Bayes Theorem Question

[Yeti]

Or at least I think so.

Here it is :



You have 5 coins coloured blue, red, white, green and yellow. One of them is unfair and comes up heads with probability 3/4. The other four coins are fair.

The blue coin is tossed 3 times and comes up head each time. What is the probability that the blue coin is unfair?



I keep getting 9/41 but I think thats wrong. I heard 4/15 off someone, don't know if that's right though.

Any help much appreciated.

[magiluke]

Ok, without any explanation, I'm going to give you .428

I'm not sure how right that is, you know how hard probablility can be. Anyway, I'm curious how you got your answer.

[willie24]

I don't know what bayes theorem states. but this is the answer i get:

(27/64)/[(1/8)(4)+(27/64)]= .4576 or 27/59

the way i did this: the crooked coin has a 27/64 chance of coming up heads 3 consecutive times. the other 4 coins each have a 1/8 chance of coming up heads each time. if you flipped all the coins 3 times in a row the same number of times, you would expect to get the all heads result 1/8 of the time for 4 of the coins, and 27/64 of the time for one coin. so then, the expectation for the crooked coin over the combined expectation for all 5 coins should give you the chance that a coin that produces the all-heads result is the crooked coin.

im not a probability wiz. this is just my own logic, so let me know if i'm wrong.

[jimdmcevoy]

Although I don't fully understand willie's explanation, I get the same answer as him, so for what it's worth, my vote is 27/59 also.

[BruceZ]

[ QUOTE ]
Or at least I think so.

Here it is :



You have 5 coins coloured blue, red, white, green and yellow. One of them is unfair and comes up heads with probability 3/4. The other four coins are fair.

The blue coin is tossed 3 times and comes up head each time. What is the probability that the blue coin is unfair?



I keep getting 9/41 but I think thats wrong. I heard 4/15 off someone, don't know if that's right though.

Any help much appreciated.

[/ QUOTE ]

P(blue unfair) = P(blue unfair AND blue 3 heads in row) / P(blue 3 heads in row)

= P(blue unfair)*P(unfair 3 heads in row) / [ P(blue unfair)*P(unfair 3 heads in row) + P(blue fair)*P(fair 3 heads in row) ]

(1/5)*(3/4)^3 / [(1/5)*(3/4)^3 + 4/5 * (1/2)^3 ]

= 27/59.

[Yeti]

Cheers guys.

I actually had the method right but had somehow worked out (3/4)^3 as 9/64 or something /images/graemlins/smile.gif

Thanks again.

[magiluke]

Mine was very close to that, maybe I miscopied something there.