# probabilities

posted by on .

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.

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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

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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...

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but Reiny it's WITH replacement!

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

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Sorry I meant 5/9

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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