probably a super easy question

[pendragon]

All these answers under 50% have to be wrong, right? That can't be right, the 10 red ball bag has to be more likely than the 5/5 bag.

This is a Bayes Theorem issue, so:

Event A = the bag is the 10 red ball bag
Event B = 2 red balls picked

P(B|A) = 1 (only red balls in this bag)
P(A) = .5 (either bag equally likely)
P(B) = 15/20*14/19 = .55263 (I'm not entirely sure about this one...it might also be .5*(1^2+.5^2)=.625)

So P(A|B)=1*.5/.55263=90.476%
Or 1*.5/.625=80% (I think this is it actually)

If I'm wrong it's probably about P(B), but I'd say the answer has to be above 50%.

Edit: Nevermind, I didn't realize that the bag was spun twice [img]/images/graemlins/smile.gif[/img] Bruce already did it.

[BruceZ]

I thought I'd rewrite this more succinctly. The notation P(A | B) means the probability that A occurs GIVEN that B occurred. It is equal to P(A and B)/P(B).

CASE A: Choose bag with all reds both times.
P(CASE A) = 1/2 * 1/2 = 1/4.

CASE B: Chose bag with reds and greens both times.
P(CASE B) = 1/2 * 1/2 = 1/4.

CASE C: Chose one of each bag.
P(CASE C) = 1/2 since the probabilities of all possible cases must sum to 1.


We want P(CASE A | both red)

= P(CASE A AND both red) / P(both red)

= P(CASE A AND both red) /
[ P(CASE A AND both red) + P(CASE B AND both red) + P(CASE C AND both red) ]

= P(CASE A)*P(both red | CASE A) /
[ P(CASE A)*P(both red | CASE A) + P(CASE B)*P(both red | CASE B) + P(CASE C)*P(both red | CASE C) ]

= (1/4 * 1) / [ (1/4 * 1) + (1/4 * 5/10 * 4/9) + (1/2 * 1/2 * 1) ]

= 0.45 or 9/20.