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11 Oct 2021, 01:30
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33% (02:29) correct
67%
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A bag contains only black red and gold balls. Any ball is equally likely to be picked. If you pick one ball at random, is it more likely to be red than black?
(1) If you remove half of the red balls, the probability of picking the black ball would be 50%
(2) If you remove the third of black balls, the probability of picking a red ball would be 60%
(1) If you remove half of the red balls, the probability of picking the black ball would be 50%
(2) If you remove the third of black balls, the probability of picking a red ball would be 60%
28 Oct 2021, 06:06
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The question says we have 3 kinds of balls- Red (R), Black (B) and Golden (G) and each ball is equally likely to be drawn.
We are asked if the probability of red ball being drawn is more than that of a black ball.
Statement 1:
It says if we remove half the number of red balls, so we are left with only half of the red balls, R/2.
If we had R/2, B, G, the prob of black ball=0.5
Black ball/total Balls= 1/2
B/(B+R/2+G)= 1/2
Cross multiply, 2B=B+R/2+G
B=R/2+G,
If R=2, G=1, B=2 => Black or red equally likely.
If R=2, G=3, B=4 => Black is more likely.
We cannot conclude if Statement 1 is sufficient.
Statement 2:
It says if we remove a third of the number of black balls, so we are left with only two-thirds of the black balls, 2B/3.
If we had R, 2B/3, G, the probability of red ball=0.6
Red ball/total Balls= 0.6
R/(2B/3+R+G)= 3/5
Cross multiply, 5R=3R+2B+3G
2R=2B+3G
R=B+1.5G
Clearly Red is more in number.
We can conclude that Statement 2 is sufficient.
We are asked if the probability of red ball being drawn is more than that of a black ball.
Statement 1:
It says if we remove half the number of red balls, so we are left with only half of the red balls, R/2.
If we had R/2, B, G, the prob of black ball=0.5
Black ball/total Balls= 1/2
B/(B+R/2+G)= 1/2
Cross multiply, 2B=B+R/2+G
B=R/2+G,
If R=2, G=1, B=2 => Black or red equally likely.
If R=2, G=3, B=4 => Black is more likely.
We cannot conclude if Statement 1 is sufficient.
Statement 2:
It says if we remove a third of the number of black balls, so we are left with only two-thirds of the black balls, 2B/3.
If we had R, 2B/3, G, the probability of red ball=0.6
Red ball/total Balls= 0.6
R/(2B/3+R+G)= 3/5
Cross multiply, 5R=3R+2B+3G
2R=2B+3G
R=B+1.5G
Clearly Red is more in number.
We can conclude that Statement 2 is sufficient.
28 Oct 2021, 10:36
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Bunuel
Breaking Down the Info:
Note that we have three possible colors. We can always replace any red/black ball for gold when our information is incomplete.
Statement 1 Alone:
We may start with the maximum red ball case. That would be 50% black and 50% red after removing. Then before removing we would have had \(Black : Red = 50\%:100\%\), hence more red than black. We could also have 1 red and let Black > Red. Then this statement is insufficient.
Statement 2 Alone:
We can start with the maximum black ball case. \(Red : Black = 60\%:40\%\) after removing.
We removed \(\frac{1}{3}\) of the black balls, which is the same as multiplying the amount of black balls by \(\frac{2}{3}\).
Then we need to multiply the number of black balls by \(\frac{3}{2}\) to return to the "before remove" state. Then \(Red : Black = 60\%:60\%\) before removal.
Recall that this was the maximum black ball case, so Red > Black for the rest of the cases. The current \(Red : Black = 60\%:60\%\) case is also a hypothetical case since we're supposed to have at least 1 gold ball. Then for all real cases, we have Red > Black, and this statement is sufficient.
Answer: B
