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 May 2nd, 2015, 02:27 AM #1 Newbie   Joined: May 2014 From: Italy Posts: 6 Thanks: 0 Moving balls between bags - stumped! Bag A contains 3 red balls and 7 green balls. Bag B contains 7 red and 8 green balls. A ball is taken out of bag A and placed in bag B. Then a ball is taken out of B. a) Calculate that both the balls are red. P(A is red) = 3/10 then, since A has to come out we add the red to bag B and it has 8 red and 8 green so the probability of the second even is P(B is red) = 8/16. So P = (3/10).(8/16) = (3/10).(1/2) = 3/20 b) Both balls are green Same sort of reasoning gives (7/10).(9/16) = 63/160 c) At least one ball is green Here I'm flummoxed. I'm missing some reasoning. d) At least one is not green. Here I'm flummoxed. I'm missing some reasoning.
 May 2nd, 2015, 06:11 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 (a) and (b) are correct. Very good! In general, if "X", "Y", "Z" are mutually exclusive events then P(X or Y or Z)= P(x)+ P(y)+ P(Z). Here we can take X to be "the first ball is red and the second is green" There are initially 3 red balls out of 10 so the probability the first ball is red is 3/10. Now there are 9 balls, 7 of which are green. The probability the second ball is green is 7/9. P(X)= (3/10)(7/9)= 7/30. Take Y to be "the first ball is green and the second is red". There are 7 green balls out of 10 so the probability the first ball is green is 7/10. Now there are 9 balls, 3 of which are red. The probability the second ball is red is 3/9. P(Y)= (7/10)(3/9)= 7/30. Take Z to be "both balls are red" As you calculated before, P(Z)= 3/20. The probability "at least one ball is not green" is 7/30+ 7/30+ 3/20= 14/60+ 14/60+ 9/20= 37/60.
May 2nd, 2015, 07:08 AM   #3
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
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Hello, ofransen!

Quote:
 Bag A contains 3 red balls and 7 green balls. Bag B contains 7 red and 8 green balls. A ball is taken out of bag A and placed in bag B. Then a ball is taken out of B. Calculate the probability that: a) Both balls are red. $\;\;\;P(\text{both red}) \:=\: \left(\frac{3}{10}\right)\left(\frac{8}{16}\right) \:=\: \frac{3}{20} \;\;\bf \text{Y\!es!}$ b) Both balls are green. $\;\;\;P(\text{both green}) \:=\:\left(\frac{7}{10}\right)\left(\frac{9}{16} \right) \:=\: \frac{63}{160}\;\;\bf\text{Right!}$

Quote:
 c) At least one ball is green.

The opposite of "at least one green" is "no green" or "both red".

$P(\text{at least one green) \;=\; 1\,-\,P(\text{both red}) \;=\;1\,-\,\frac{3}{20} \;=\;\frac{17}{20}$

Quote:
 d) At least one is not green.

The opposite of "At least one not green" = "Both green"

$P(\text{At least one not green}) \;=\; 1\,-\,P(\text{both green}) \;=\;1\,-\,\frac{63}{160} \;=\;\frac{97}{160}$

 May 2nd, 2015, 11:48 PM #4 Newbie   Joined: May 2014 From: Italy Posts: 6 Thanks: 0 Oh it's so easy when explained! I knew I was missing something...

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