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A couple probability questions

  1. Mar 19, 2009 #1
    1. An urn contains 2 white balls and 8 red balls. A second urn contains 8 white balls and 2 red balls. An urn is selected, and the probability of selecting the first urn is 0.4. A ball is drawn from the selected urn and replaced. Then another ball is drawn and replaced from the same urn. If both balls are white, what are the following probabilities? (Round your answers to three decimal places.)

    (a) the probability that the urn selected was the first one

    I tried multiplying the probability of the first urn .4 x the probability of a white ball .2 together, then squaring it to get the probability of it happening twice. Clearly not correct though. Do I have to use bayes theorem?

    (b) the probability that the urn selected was the second one
    This I know is 1-the answer to a


    2. A box has 4 blue and 2 green jelly beans. A bag has 8 blue and 6 green jelly beans. A jelly bean is selected at random from the box and placed in the bag. Then a jelly bean is selected at random from the bag. If a green jelly bean is selected from the bag, what is the probability that the transferred jelly bean was green? (Round your answer to three decimal places.)

    I figured out the probability of picking a green jelly bean from the bag, and the probability of picking the other options from the box. And I know we are supposed to find the probability of picking a green from the box given picking a green from the bag. However I can not get it to work in the equations that I have.
     
  2. jcsd
  3. Mar 20, 2009 #2
    you can use Baye's theorem..I just used a probability tree. But, it doesn't change anything. The probability of picking the first urn is .4, as stated. Kind of like the Monty Hall problem, the probability doesn't change

    Second, one using probability tree, brute force, probability is 7/19. Sorry can't help with equations.
     
  4. Mar 21, 2009 #3
    Thanks for the help. I tried using bayes theorem for the first problem by doing the probability of the first earn given the ball is white. So 1/W. Which is .4x.2/(.4x.2 + .6x.8) and got .1429. Then multiplied that by .4x.2 because knowing it is the same earn and the ball is white, the probability of picking another white ball is .08. But this was not correct. Where did I go wrong? -Thanks
     
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