help with math problem

[furyshade]

this was given as extra credit, but no one in any of the classes this teacher has (highschool) got it that i know of, so wondering if we were missing something very obvious.

You have two hats, hat #1 has 8 red marbles and 2 blue marbles, hat #2 has 2 red marbles and 8 blue. you randomly select a hat and randomly choose two marbles from it, both of which are blue. you then switch hats and randomly pick out a marble, what is the probability that this will be a red marble?

usually we are not allowed calculators on these extra credit problems but for this one it said a calculator was allowed and to round to the third decimal place.

[tabako]

You need to use bayes' theorem to find the probability that you selected the two blue marbles from a given hat, and then from there it is straight forward to calculate the chance that you grab a red marble from the other hat.

[furyshade]

[ QUOTE ]
You need to use bayes' theorem to find the probability that you selected the two blue marbles from a given hat, and then from there it is straight forward to calculate the chance that you grab a red marble from the other hat.

[/ QUOTE ]
i dont think any of us have heard of baye's theorem, and i dont get how the first picking affects the second

[qwnu]

[ QUOTE ]
[ QUOTE ]
You need to use bayes' theorem to find the probability that you selected the two blue marbles from a given hat, and then from there it is straight forward to calculate the chance that you grab a red marble from the other hat.

[/ QUOTE ]
i dont think any of us have heard of baye's theorem, and i dont get how the first picking affects the second

[/ QUOTE ]

Once you pick a hat, all you know is that there's a 50% chance you have hat A, and a 50% chance you have hat B.

But after you pick out two blue marbles from your first hat, you have some additional information. It should be intuitively obvious that it is now more likely that you have hat B. Bayes Theorem allows you to calculate this likelihood precisely.

[Sephus]

113/145

[TomCowley]

113/145 is what I got, forum ate my post and I'm not writing it again.

[furyshade]

i know nothing about statistics/probability other than the most basic stuff, is there any simple way to go about it or does it simply require knowledge i don't have at the moment?

[kevin017]

this is definitely doable with high school math. i've never heard of bayes theorem, and i just did probably the most convulted and horrible process of solving it, but i came up with 77.9%, or 113/145.

Ok, i just read your above post, i will include what it is that I did. this may be horrible, but here it is.

The odds of drawing two blue marbles from hat 1 are 2/10, then 1/9, or multiply them together to get 2.2%.

The odds of drawing two blue marbles from hat 2 are 8/10, then 7/9, or multiply them together to get 62.2%.

So, we want to set this to 100%. so we add them together, and get 64.4 (62.2 + 2.2), then we divide. 2.2/64.4 = 3.4, and 62.2 / 64.4 = 96.6.

So, the odds of us having gotten the first two blue marbles from hat a is 3.4%, and the odds of us having gotten them from hat b is 96.6%.

So if the first hat was A, the second is B, and the odds of drawing a red marble from b is 2/10. 3.4% * 20% = .68%

If the first hat was B, the second is A, and the odds of drawing a red marble from A is 8/10. 96.6% * 80% = 77.28%

77.28% + .68% = 77.96%

[furyshade]

[ QUOTE ]
this is definitely doable with high school math. i've never heard of bayes theorem, and i just did probably the most convulted and horrible process of solving it, but i came up with 77.9%, or 113/145.

[/ QUOTE ]

id like to hear it, im generally pretty good at math, this extra credit was given to the three highest math classes at my school, though none of them are statistics courses, no one got it that i know of.

i tried reading the wiki on baye's theorem but it gave a lot of notation that i haven't seen

[kevin017]

i'm hoping that what i wrote is right. I think it is because I came up with the same answer as others who are probably right, though i could be totally wrong, and i probably went about doing it in a weird way.