Question

Probability Help!!!

An urn contains 10 balls: 4 red and 6 blue. A second urn contains 16 red balls and an unknown number of blue balls. A single ball is drawn from each urn. The probability that both balls are the same color is .44. Calculate the number of blue balls in the second urn.

Answer 1

HI!!

Answer 2

Hi :-)

Answer 3

ok i read it wrong
there are 16 red balls, not 16 total

Answer 4

this is a conditional probability problem isn't it?

Answer 5

@dan815

Answer 6

@SolomonZelman

Answer 7

this seems, interesting... haha gimme a sec to try it

Answer 8

okay thank you

Answer 9

Because I mean...it seems easy enough to break down...we know if it is either R,R or B,B that probability is .44

\[\large P((R_1and R_2)\cup (B_1 and B_2)) = .44 \]

So break that up into

\[\large P(R_1 and R2) + P(B_1 and B_2)\]

Meaning

\[\large P(R_1)\times P(R_2) + P(B_1) \times P(B_2)\]

\[\large \frac{4}{10} \times \frac{16}{x + 16} + \frac{6}{10} \times \frac{x}{x + 16}\]

Then just solve for 'x'

Answer 10

Well...I should have brought down the .44 after that...but yeah that = .44 and solve for 'x'

Answer 11

okay i will solve and see... i guess when i think of the union of those two events... i also thought you have to subtract the intersection as well.. but the intersection of those two events can't happen

Answer 12

Wow apparently I cannot solve a simple algebra problem, having trouble finding 'x' XD

Answer 13

Oh no, nvm lol divided instead of multiplied haha, damn mistakes

Answer 14

ya i got x = 4, which is what the answer is. i am still confused though on the initial formula... because i think of
A union B = A + B - A intersect B

Answer 15

Well the question states, if 2 of the same color ball are chosen, that probability is .44

Which occurs when (R,R) or (B,B) happens

I know that you're thinking of \(\large (A \cup B) = P(A) + P(B) - P(A or B)\)

But that occurs when we have 2 events and want to know if either, or neither occurs, however here there is no intersection in this problem

Answer 16

I just found an example where you would use that, and it should make this whole thing make sense to you

A card is drawn randomly from a deck of ordinary playing cards. You win $10 if the card is a spade or an ace. What is the probability that you will win the game?

P(spade) = 13/52
P(ace) = 4/52

BUT

P(spade AND ace) = intersection = 1/52

Answer 17

That is a problem where you would use that rule

however here, we just want an event to occur GIVEN that another event has already occured

Answer 18

Ya and thats where i was getting confused because i was trying to use conditional probability and P(R2 given R1) and so on... but this makes sense... i appreciate the help and hope you don't mind if i get back at you with some more probability sometime.

Answer 19

No problem, I took stats almost 2 years ago so Im a little rusty but I'll try and help when I can lol

Answer 20

thank you