Mycroft has a bag that only contains few red, blue, and green nuggets of identical shape, size, and weight. Two nuggets are randomly drawn from the bag, without replacement. What is the difference in the probability that both nuggets so drawn are red and the probability that none is green?
(1) Mycroft has ¾ as many red nuggets as the blue nuggets in his bag, which contains 1 green nugget more than the red nuggets.
(2) The probability that "either nuggets so drawn are red or none is green" is 53/105.
Made Up!
the difference in the probability
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- sanju09
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R, B, G --> 2 are drawn
To find: Prob of Red -(Prob of NOT Green) ==> [(R)(R-1) - (R+B)(R+B-1)]/(R+B+G)(R+B+G-1)
Statement 1:
3/4 R = B ==> 3R = 4B
G = R + 1
If B = 3, then R = 4 & G = 5 ==> Probability = |(4x3 - 7x6)|/(12)(11) = 30/12*11 = 5/22
If B = 6, then R = 8 & G = 9 ==> probability = | 8x7 - 14x13|/(23)(22) = 126/23*22 = 63/263
INSUFFICIENT
Statement 2:
((R)(R-1) + (R+B)(R+B-1))/(R+B+G)(R+B+G-1) = 53/120
we cannot conclude on separate probabilities
INSUFFICIENT
Combining...
((R)(R-1) + (R+B)(R+B-1))/(R+B+G)(R+B+G-1) = 53/120
we can convert the equation say in terms of "R" using:
3/4 R = B
G = R + 1
IF we get R then we can get all other values
SUFFICIENT
Answer [spoiler]{C}[/spoiler]?
To find: Prob of Red -(Prob of NOT Green) ==> [(R)(R-1) - (R+B)(R+B-1)]/(R+B+G)(R+B+G-1)
Statement 1:
3/4 R = B ==> 3R = 4B
G = R + 1
If B = 3, then R = 4 & G = 5 ==> Probability = |(4x3 - 7x6)|/(12)(11) = 30/12*11 = 5/22
If B = 6, then R = 8 & G = 9 ==> probability = | 8x7 - 14x13|/(23)(22) = 126/23*22 = 63/263
INSUFFICIENT
Statement 2:
((R)(R-1) + (R+B)(R+B-1))/(R+B+G)(R+B+G-1) = 53/120
we cannot conclude on separate probabilities
INSUFFICIENT
Combining...
((R)(R-1) + (R+B)(R+B-1))/(R+B+G)(R+B+G-1) = 53/120
we can convert the equation say in terms of "R" using:
3/4 R = B
G = R + 1
IF we get R then we can get all other values
SUFFICIENT
Answer [spoiler]{C}[/spoiler]?
R A H U L
- sanju09
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Nice thought process Rahul. But you must thank it was DS, had it been PS, you'd have paid heavily for reading 105 as 120.theCodeToGMAT wrote:R, B, G --> 2 are drawn
To find: Prob of Red -(Prob of NOT Green) ==> [(R)(R-1) - (R+B)(R+B-1)]/(R+B+G)(R+B+G-1)
Statement 1:
3/4 R = B ==> 3R = 4B
G = R + 1
If B = 3, then R = 4 & G = 5 ==> Probability = |(4x3 - 7x6)|/(12)(11) = 30/12*11 = 5/22
If B = 6, then R = 8 & G = 9 ==> probability = | 8x7 - 14x13|/(23)(22) = 126/23*22 = 63/263
INSUFFICIENT
Statement 2:
((R)(R-1) + (R+B)(R+B-1))/(R+B+G)(R+B+G-1) = 53/120
we cannot conclude on separate probabilities
INSUFFICIENT
Combining...
((R)(R-1) + (R+B)(R+B-1))/(R+B+G)(R+B+G-1) = 53/120
we can convert the equation say in terms of "R" using:
3/4 R = B
G = R + 1
IF we get R then we can get all other values
SUFFICIENT
Answer [spoiler]{C}[/spoiler]?
Best
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
- theCodeToGMAT
- Legendary Member
- Posts: 1556
- Joined: Tue Aug 14, 2012 11:18 pm
- Thanked: 448 times
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- GMAT Score:650
ooops......sanju09 wrote:
Nice thought process Rahul. But you must thank it was DS, had it been PS, you'd have paid heavily for reading 105 as 120.
Best
R A H U L