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December 22, 2014

December 22, 2014

Posted by **joanna** on Tuesday, November 20, 2012 at 7:27am.

- probabilities -
**Reiny**, Tuesday, November 20, 2012 at 8:00amHe 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 -
**Mathtaculator**, Tuesday, November 20, 2012 at 8:00ammake 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 -
**joanna**, Tuesday, November 20, 2012 at 8:03ambut Reiny it's WITH replacement!

- probabilities -
**joanna**, Tuesday, November 20, 2012 at 8:04amI got 4/5 but the answer in the book says something else. Am I correct?

- probabilities -
**joanna**, Tuesday, November 20, 2012 at 8:10amSorry I meant 5/9

- probabilities -
**Reiny**, Tuesday, November 20, 2012 at 8:11amRight! , 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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