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anonymous
 one year ago
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.
anonymous
 one year ago
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.

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misty1212
 one year ago
Best ResponseYou've already chosen the best response.0ok i read it wrong there are 16 red balls, not 16 total

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is a conditional probability problem isn't it?

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1this seems, interesting... haha gimme a sec to try it

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1Because 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'

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1Well...I should have brought down the .44 after that...but yeah that = .44 and solve for 'x'

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay 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

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1Wow apparently I cannot solve a simple algebra problem, having trouble finding 'x' XD

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1Oh no, nvm lol divided instead of multiplied haha, damn mistakes

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ya 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

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1Well 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

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1I 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

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1That 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ya 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.

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.1No problem, I took stats almost 2 years ago so Im a little rusty but I'll try and help when I can lol
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