Simple problems thread

>>9108196
one of them contains 100 red chips*

>>9108196
At one time I probably would've been able to solve this.

Is it the probability of getting ten red chips out of the bag with blue chips times the probability of choosing the bag with blue chips (50%)?

so like 0.02967%?? for the first question

>>9108229

yeah, first one is easy, just use the F.C.P. and multiply your events. Basically you have 11 events. The first is the .5 chance you even pick the right bag, and then you'll have the probability of pulling out a red chip each successive turn. Easy.

For the second one just multiply (.5) x (50/90), since there's still a 50% chance you have the right bag and then just multiply the P(blue) if you do have the right bag

>>9108238
right, thanks

that goat thing with the three doors destroyed all my confidence concerning these types of questions lol

>>9108238
After drawing 10 reds in a row the chance you have the blues bag is significantly less than 50%. You have to update that discrete uniform assumption like this

P(bag with blue | draw 10 reds) = P(bag with blue and draw 10 reds) / P(draw 10 reds) = P(draw 10 reds | bag with blue) * P(bag with blue (this is the .5 probability)) / P(draw 10 reds)

Whichever one is easier to solve. It makes more intuitive sense to update the probability too. P(not bag with blue | 10r ) = 1 - the one above obviously.

>>9108344
And that's called bayesian updating, if you were doing any serious application you would take all information into account to adjust your "prior" distributions. First you choose 50 50 between the two bags but as you draw more and more reds you're thinking the chance you picked the bag with no blues is higher.
The general form is
New distribution given some information = Probability of information given old distribution times old distribution divided by the information given old distribution integrated/summed over all possibilities of old (a normalizing constant)

Formally "new" is called posterior and "old" is prior