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March 30, 2017

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A bag contains 3 green balls. 2 red balls and a white ball. A boy randomly draws balls from the bag one at a time (with replacement) until a red ball appears. Find the probability that he will make at least 3 draws.

  • probabilities - ,

    He could get it on
    the first draw
    or the 2nd draw
    or the 3rd draw
    or the 4th draw
    or the 5th draw
    or the 6th draw --- , at this point he must have picked it

    = 2/6 + (4/6)(2/5) + (4/6)(3/5)(2/4) + (4/6)(3/5)(2/4)(2/3) + (4/6)(3/5)(2/4)(1/3)(2/2) + 0
    = 1/3 + 4/15 + 1/5 + 2/15 + 1/15
    notice that this adds up to 1

    so prob of at least 3 draws = 1 - (1/3 + 4/15) = 2/5

  • probabilities - ,

    make a tree chart. one scenario would be drawing a red ball, another a white ball, another...etc. Then, after you pick one case, make other choices... like after i picked a red ball...i picked either a red ball again or a white ball...

  • probabilities - ,

    but Reiny it's WITH replacement!

  • probabilities - ,

    I got 4/5 but the answer in the book says something else. Am I correct?

  • probabilities - ,

    Sorry I meant 5/9

  • probabilities - ,

    Right! , should read the question more carefully

    so my string would be
    (2/6) + (4/6)(2/6) + (4/6)(4/6)(2/6) + (4/6)(4/6)(4/6)(2/6) + .....
    = (1/3) + (2/3)(1/3) + (2/3)^2 (1/3) + (2/3)^3 (1/3) + ...
    This must have a total of 1

    so the prob of least 3 draws
    = 1 - (1/3 + 2/9) = 4/9

    note my string of additions is a geometric series
    where a = 1/3 and r = 2/3
    sum(∞) = a/(1-r)
    = (1/3) / (1-2/3) = 1

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