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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.

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  1. misty1212
    • one year ago
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    HI!!

  2. anonymous
    • one year ago
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    Hi :-)

  3. misty1212
    • one year ago
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    ok i read it wrong there are 16 red balls, not 16 total

  4. anonymous
    • one year ago
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    this is a conditional probability problem isn't it?

  5. anonymous
    • one year ago
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    @dan815

  6. anonymous
    • one year ago
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    @SolomonZelman

  7. johnweldon1993
    • one year ago
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    this seems, interesting... haha gimme a sec to try it

  8. anonymous
    • one year ago
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    okay thank you

  9. johnweldon1993
    • one year ago
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    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'

  10. johnweldon1993
    • one year ago
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    Well...I should have brought down the .44 after that...but yeah that = .44 and solve for 'x'

  11. anonymous
    • one year ago
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    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

  12. johnweldon1993
    • one year ago
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    Wow apparently I cannot solve a simple algebra problem, having trouble finding 'x' XD

  13. johnweldon1993
    • one year ago
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    Oh no, nvm lol divided instead of multiplied haha, damn mistakes

  14. anonymous
    • one year ago
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    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

  15. johnweldon1993
    • one year ago
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    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

  16. johnweldon1993
    • one year ago
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    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

  17. johnweldon1993
    • one year ago
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    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

  18. anonymous
    • one year ago
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    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.

  19. johnweldon1993
    • one year ago
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    No problem, I took stats almost 2 years ago so Im a little rusty but I'll try and help when I can lol

  20. anonymous
    • one year ago
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    thank you

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