A bag contains 12 balls some of which are red and rest are blue. How many red balls are there in the bag?
1) The probability of randomly selecting two red balls from the bag is 1/11.
2) When two balls are selected randomly from the bag the probability of selecting one red and one blue ball is 9/22
ds:probability
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- HSPA
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using 1 : 4/12 * 3/11 will do => 4 red
x, 12-x
(x/12 + 12-x-1/11) + 8/12 * 4/11.... got to go havent calculated this..
I guess D
x, 12-x
(x/12 + 12-x-1/11) + 8/12 * 4/11.... got to go havent calculated this..
I guess D
First take: 640 (50M, 27V) - RC needs 300% improvement
Second take: coming soon..
Regards,
HSPA.
Second take: coming soon..
Regards,
HSPA.
- bubbliiiiiiii
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IMO D.
Let the number of red balls be x => blue are 12-x. we have to find x.
Option 1:
xc2/12c2=1/11 : Solve for X SUFFICIENT
Option 2:
(xc1*(12-x)c1)/12c2=9/22 : Solve for X SUFFICIENT
Let the number of red balls be x => blue are 12-x. we have to find x.
Option 1:
xc2/12c2=1/11 : Solve for X SUFFICIENT
Option 2:
(xc1*(12-x)c1)/12c2=9/22 : Solve for X SUFFICIENT
Regards,
Pranay
Pranay
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Careful! Let's think about statement 2 again. Logically, the "x" in your equation could just as easily refer to the blue balls rather than red. Although we would have two integers whose sum was 12, we wouldn't know which integer referred to which color ball.bubbliiiiiiii wrote:IMO D.
Let the number of red balls be x => blue are 12-x. we have to find x.
Option 1:
xc2/12c2=1/11 : Solve for X SUFFICIENT
Option 2:
(xc1*(12-x)c1)/12c2=9/22 : Solve for X SUFFICIENT
You can also see this algebraically. We know that xC1 is always x. So, if you sub in "x" for "xc1" and "12-x" for "(12-x)c1", you will end up with a quadratic equation, which of course will have two solutions.
The correct answer is A.
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- bubbliiiiiiii
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Thanks testluv a great concept shared.Testluv wrote:Careful! Let's think about statement 2 again. Logically, the "x" in your equation could just as easily refer to the blue balls rather than red. Although we would have two integers whose sum was 12, we wouldn't know which integer referred to which color ball.bubbliiiiiiii wrote:IMO D.
Let the number of red balls be x => blue are 12-x. we have to find x.
Option 1:
xc2/12c2=1/11 : Solve for X SUFFICIENT
Option 2:
(xc1*(12-x)c1)/12c2=9/22 : Solve for X SUFFICIENT
You can also see this algebraically. We know that xC1 is always x. So, if you sub in "x" for "xc1" and "12-x" for "(12-x)c1", you will end up with a quadratic equation, which of course will have two solutions.
The correct answer is A.
Regards,
Pranay
Pranay