Two marbles are drawn from a jar with 10 marbles. If all marbles are either red or blue, is the probability that both marbles selected will be red greater than 3/5?
(1) The probability that both marbles selected will be blue is less than 1/10.
(2) At least 60% of the marbles in the jar are red.
OA: E
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Two marbles are drawn from a jar with 10 marbles. If all mar
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richachampion
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Last edited by richachampion on Tue Sep 13, 2016 5:47 am, edited 1 time in total.
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joealam1
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Hi Richampion,
Nice one! but to me E is the answer, i couldn't figure out how C cloud be correct; below is my solution:
is P(both red) > 3/5 ?
=> P( both red)= R/10 * (R-1)/9 > 54/90?
1)Insufficient:
B*(B-1)/90 <1/10
=> B(B-1)/9 < 1 => B(B-1)<9 so B can be {3,2 or 1}
if B=3 then R=7 then P(both red)= 7*6/90 = 42/90 < 54/90 (No)
if B=1 then R=9 then P(both red)= 9*8/90 = 72/90 > 54/ 90 (Yes)
2)Insufficient:
At least 60% are red:
=> if Red are 70% then R=7 then then R=7 then P(both red)= 7*6/90 = 42/90 < 54/90 (No)
=> if Red are 90% then R=9 then P(both red)= 9*8/90 = 72/90 > 54/ 90 (Yes)
Combining both 1) and 2)
from 2) at least 60% are red, so let's say 70% are red then R=7 and B=3
with B=3 P(both B)= 3*2/90 <1/10 satisfying condition 1), however P(both R)= 7*6/90= 42/90 < 54/90 (No)
if 90% are red then R=9 and B=1
P(both B)= 1*0/90= 0 <1/10 satisfying condition 1)
P(both R)= 9*8/90= 72/90 > 54/90 (Yes)
what's the source of this exercise??
Nice one! but to me E is the answer, i couldn't figure out how C cloud be correct; below is my solution:
is P(both red) > 3/5 ?
=> P( both red)= R/10 * (R-1)/9 > 54/90?
1)Insufficient:
B*(B-1)/90 <1/10
=> B(B-1)/9 < 1 => B(B-1)<9 so B can be {3,2 or 1}
if B=3 then R=7 then P(both red)= 7*6/90 = 42/90 < 54/90 (No)
if B=1 then R=9 then P(both red)= 9*8/90 = 72/90 > 54/ 90 (Yes)
2)Insufficient:
At least 60% are red:
=> if Red are 70% then R=7 then then R=7 then P(both red)= 7*6/90 = 42/90 < 54/90 (No)
=> if Red are 90% then R=9 then P(both red)= 9*8/90 = 72/90 > 54/ 90 (Yes)
Combining both 1) and 2)
from 2) at least 60% are red, so let's say 70% are red then R=7 and B=3
with B=3 P(both B)= 3*2/90 <1/10 satisfying condition 1), however P(both R)= 7*6/90= 42/90 < 54/90 (No)
if 90% are red then R=9 and B=1
P(both B)= 1*0/90= 0 <1/10 satisfying condition 1)
P(both R)= 9*8/90= 72/90 > 54/90 (Yes)
what's the source of this exercise??
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richachampion
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R/10 X (R-1)/9 > 3/5
R (R-1) > 9 X 10 X 3 /5 = 54
R² - R - 54 > 0
R =7 → This gives -ve Value
R = 8 → This gives +ve value.
Thus R is between 7 and 8
R > 7 = Red Balls should be greater than 7.
R can be 8, 9, or 10.
(1) The probability that both marbles selected will be blue is less than 1/10.
B/10 X (B-1)/9 < 1/10
B(B-1) < 9
B² - B - 9 < 0
By Hit and trial we know that this could be true for B =3, 2, and 1.
That means B < 4 or R > 6 (7, 8, 9, and 10) → YES , and NO situation as R can only be 8, 9, or 10.
(2) At least 60% of the marbles in the jar are red.
This statement is similar to statement 1.
There are 10 balls, thus, %ages(of probability) will increase or decrease in the multiples of 10.
Red balls will be 70% or 80% or 90% or 100% ≡ 7 or 8 or 9 or 10.
Thus both the statement eventually gives the same information. That means that their combination won't yield anything new. Thus,
E
R (R-1) > 9 X 10 X 3 /5 = 54
R² - R - 54 > 0
R =7 → This gives -ve Value
R = 8 → This gives +ve value.
Thus R is between 7 and 8
R > 7 = Red Balls should be greater than 7.
R can be 8, 9, or 10.
(1) The probability that both marbles selected will be blue is less than 1/10.
B/10 X (B-1)/9 < 1/10
B(B-1) < 9
B² - B - 9 < 0
By Hit and trial we know that this could be true for B =3, 2, and 1.
That means B < 4 or R > 6 (7, 8, 9, and 10) → YES , and NO situation as R can only be 8, 9, or 10.
(2) At least 60% of the marbles in the jar are red.
This statement is similar to statement 1.
There are 10 balls, thus, %ages(of probability) will increase or decrease in the multiples of 10.
Red balls will be 70% or 80% or 90% or 100% ≡ 7 or 8 or 9 or 10.
Thus both the statement eventually gives the same information. That means that their combination won't yield anything new. Thus,
E
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richachampion
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I posted E.joealam1 wrote:Exactly!! but didn't you post that the OA is C??
what's the source of this exercise??
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Matt@VeritasPrep
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Let's say we have r red and b blue. The probability of getting both red would be
r/10 * (r - 1)/9, or (r² - r)/90
If this is greater than 3/5, we have
(r² - r)/90 > 3/5, or
5r² - 5r > 270, or
r² - r > 54
Since r is a nonnegative integer, this is the same as "Is r > 8?"
S1:
b/10 * (b - 1)/9 < 1/10
Using the same simplification process as above, we get b < 4. This doesn't QUITE answer our question, since we could have b = 3, r = 7 or b = 1, r = 9; NOT SUFFICIENT.
S2:
r ≥ 6. Not helpful at all, S1 already told us this; NOT SUFFICIENT.
S1 + S2:
Nothing new! S2 implies S1, so we're stuck, and the answer is E.
r/10 * (r - 1)/9, or (r² - r)/90
If this is greater than 3/5, we have
(r² - r)/90 > 3/5, or
5r² - 5r > 270, or
r² - r > 54
Since r is a nonnegative integer, this is the same as "Is r > 8?"
S1:
b/10 * (b - 1)/9 < 1/10
Using the same simplification process as above, we get b < 4. This doesn't QUITE answer our question, since we could have b = 3, r = 7 or b = 1, r = 9; NOT SUFFICIENT.
S2:
r ≥ 6. Not helpful at all, S1 already told us this; NOT SUFFICIENT.
S1 + S2:
Nothing new! S2 implies S1, so we're stuck, and the answer is E.
These probability questions with an inequality (more than x or less than x) are a lot harder than the ones where you just find the probability. Is there a general strategy you can use? Also what level question is this?
