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Re: One bag contains 4 white balls and 2 black balls, another bag contains [#permalink]
stolyar wrote:
a. agree
b. agree
c. P=P(BW)+P(WB)=4/6*5/8+5/8*4/6=10/12=5/6



Quote:
4/6*5/8+5/8*4/6


where you get the bold part from...

i edited my solution...see if its ok.

thanks
praetorian
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Re: One bag contains 4 white balls and 2 black balls, another bag contains [#permalink]
A. 4/6 * 3/8 = 12/48 = 1/4 = .25
B. 2/6 * 5/8 = 10/48 = 5/24 = .21
C. (4/6 * 5/8) + (2/6 * 3/8) = 20/48 + 6/48 = 26/48 = 13/24 =.54
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Re: One bag contains 4 white balls and 2 black balls, another bag contains [#permalink]
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Re: One bag contains 4 white balls and 2 black balls, another bag contains [#permalink]
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