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# M09 Q11

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Intern
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19 Nov 2008, 07:03
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The bowl contains green and blue chips. What is the probability of drawing a blue chip in two successive trials if the chip drawn in the first trial is not returned to the bowl before the second trial?

1. The ratio of blue chips to green chips is 3:4
2. There are 5 more green chips than blue chips

REVISED VERSION OF THIS QUESTION IS HERE: m09-q11-73062.html#p1212687

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[Reveal] Spoiler:
soln::
-----
S1 is not sufficient. Although we know the probability that the first chip will be blue, we cannot compute the probability that the second chip will be blue. We need to evaluate the ratio of blue chips to green chips after the first trial and S1 does not supply this information.

S2 is not sufficient either. We don't even know if this difference of 5 is significant or not.

From S1 and S2 taken together the exact number of green and blue chips in the bowl can be determined:

From this system and . This information completely defines the contents of the bowl and thus the question can be answered.

-------
but from s1. we already know it is in the ratio of 3:4 -
So, first blue chip is 3/7 and second one is in the ratio of 2/6.. right ??

why do we need to figure out the exact number ?

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20 Nov 2008, 08:56
10
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From S1 we only know the ratio of blue chips to green chips. We don't know the exact number of respective chips. The number of blue and green chips can vary greatly as long as the ratio requirement is met:
3 blue and 4 green
6 blue and 8 green
30 blue and 40 green
...

Even if the probability of drawing a blue chip during the first trial is the same for all these examples, the probability of doing the same during the second trial is different for each of the examples:
$$\frac{2}{6}$$

$$\frac{5}{13}$$

$$\frac{29}{69}$$

...

Hope this helps.
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07 Apr 2010, 11:20
Completely agree with dzyubam.
The absolute values with matter here, especially with the second trial.

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07 Apr 2010, 11:35
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x -> nr of blue, y -> nr of green

x/y = 3/4 & y-x = 5 =>
y/x = 4/3 => (y-x)/x = (4-3)/3 => 5/x = 1/3 => x = 15 => y = 20

P1(b) = 15/35 = 3/7
P2(b) = 14/34 = 7/17

P1,2(b) = 3/7∙7/17 = 3/17

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10 Apr 2010, 11:17
Great clarification dzyubam. My first reaction was the same as mbaobsessed. i figured that if the probability of the first event was 3/7, then next would be 2/6. Your example is very helpful to illustrate why you can't solve A alone. Thanks.
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11 Apr 2011, 11:05
ykpgal wrote:
x -> nr of blue, y -> nr of green

x/y = 3/4 & y-x = 5 =>
y/x = 4/3 => (y-x)/x = (4-3)/3 => 5/x = 1/3 => x = 15 => y = 20

P1(b) = 15/35 = 3/7
P2(b) = 14/34 = 7/17

P1,2(b) = 3/7∙7/17 = 3/17

x -> nr of blue, y -> nr of green
x/y = 3/4 & y-x = 5 =>
y/x = 4/3 => (y-x)/x = (4-3)/3 => 5/x = 1/3 => x = 15 => y = 20

For consecutive ratios, if we just multiply the ratios with the difference in actual values, we will get the actual value....

Eg. Ratio is 3:4 and the difference is 5. Just multiply the ratios by the difference. 3x5 and 4x5 => Actual values are 15 and 20.

Ratio 6:5, and the difference in the actual values is say 7, the values are 42 and 35

Last edited by mniyer on 11 Apr 2011, 19:03, edited 1 time in total.

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11 Apr 2011, 19:02
1
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As per (1)

Blue chips = 3x

Green chips = 4x

Probability = 3x/7x * (3x - 1)/(7x - 1)

Not Sufficient.

As per (2)

G = B + 5

Probability = B/(2B + 5) * (B-1)/(2B + 4)

Not Sufficient.

(1) and (2)

4x = 3x + 5

=> x = 5

So Probability = 3/7 * 14/34

= 3/17

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15 Apr 2011, 16:14
With out knowing number of blue chips, we cannot calculate probability of drawing a blue chip.

1. Not sufficient as we don't know number of blue chips.

2.Not sufficient as we don't know number of blue chips

Together, sufficient enough to find number of blue chips and drawing blue chip probability .

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29 Jun 2011, 03:20
subhashghosh wrote:
As per (1)

Blue chips = 3x

Green chips = 4x

Probability = 3x/7x * (3x - 1)/(7x - 1)

Not Sufficient.

As per (2)

G = B + 5

Probability = B/(2B + 5) * (B-1)/(2B + 4)

Not Sufficient.

(1) and (2)

4x = 3x + 5

=> x = 5

So Probability = 3/7 * 14/34

= 3/17

Subaash, I liked your approach. Thanks

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05 Sep 2011, 01:45
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Good concept .. Thanks everyone

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13 Apr 2012, 06:44
mbaobsessed wrote:
The bowl contains green and blue chips. What is the probability of drawing a blue chip in two successive trials if the chip drawn in the first trial is not returned to the bowl before the second trial?

1. The ratio of blue chips to green chips is 3:4
2. There are 5 more green chips than blue chips

[Reveal] Spoiler: OA
C

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soln::
-----
S1 is not sufficient. Although we know the probability that the first chip will be blue, we cannot compute the probability that the second chip will be blue. We need to evaluate the ratio of blue chips to green chips after the first trial and S1 does not supply this information.

S2 is not sufficient either. We don't even know if this difference of 5 is significant or not.

From S1 and S2 taken together the exact number of green and blue chips in the bowl can be determined:

From this system and . This information completely defines the contents of the bowl and thus the question can be answered.

-------
but from s1. we already know it is in the ratio of 3:4 -
So, first blue chip is 3/7 and second one is in the ratio of 2/6.. right ??

why do we need to figure out the exact number ?

first time, i understood something of a probabality problem and the answer is C
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13 Apr 2012, 12:45
C

1)Ratios are never really gud enough to find probability. Check with some values that are in ratio 3:4 . A is out
2) In this we have B and G = B+5 we can deduce the probability in terms of B. But we dont know the value of B

use B/B+5 = 3/4

and u r done

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16 Apr 2013, 05:24
BELOW IS REVISED VERSION OF THIS QUESTION:

A bowl contains green and blue chips only. If two chips are drawn from the bowl (without replacement) what is the probability that both chips will be blue?

(1) The ratio of blue chips to green chips is 3:4 --> if $$B=3$$ and $$G=4$$ then $$P=\frac{3}{7}*\frac{2}{6}$$ but if $$B=6$$ and $$G=8$$ then $$P=\frac{6}{15}*\frac{5}{13}$$. Two different answers, hence not sufficient.

(2) There are 5 more green chips than blue chips --> $$G=B+5$$. Clearly insufficient.

(1)+(2) B:G=3:4 and $$G=B+5$$ --> $$B=15$$ and $$G=20$$. Sufficient.

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16 Apr 2013, 07:16
Original question is not clear -- question may ask either "drawing at least one blue chip" or "drawing two blue chips" in two successive draws. However, for both the cases, logic will be the same and answer also will be the same as it is DS question.

1. To find the probability, we need the number of chips; only proportion does not give the answer.
Insufficient.

2. Number of chips is not known.
Insufficient.

1 + 2: Number of blue chips = 15 and number of green chips = 20.
--> It is possible to find the probability.
Sufficient.

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16 Apr 2013, 09:36
Niether 1 nor 2 alone is sufficient...
Combining 1 & 2:

B/G=3/4
G=B+5
2 equations, 2 variables...sufficient..Hence C is the correct answer.

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16 Apr 2013, 10:15
3
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Deceptive little question bugger
At the first look, one might feel that the knowledge of the ratio of the number of Green to Blue chips is sufficient to calculate the probability that is being asked. Let's go down this rabbit hole for a second...

Blue Chips : Green Chips = 3 : 4
Lets convert the ratios into absolutes.
Let # of Blue Chips = 3x
Let # of Green Chips = 4x
Total # of Chips = 7x

Probability of picking up 2 Blue chips in successive attempts = P (Blue in first attempt) * P(Blue in second attempt ).

[*]Probability (Blue in first attempt) = 3x/7x
[*]After the first attempt, the total # of blue chips = 3x - 1
[*]After the first attempt, the total # of chips = 7x - 1

Probability of picking up blue chip in the second attempt : ( 3x-1)/(7x-1)
Required probability of picking up two successive blue chips = (3x/7x) * (3x-1)/(7x-1)

No matter what we do, we cannot eliminate the "x".
Hence (A) is not sufficient

The key take away from this question is that, given the ratio of the number of Green : Blue chips, , we can always find out the probability of picking either colored chip in the first attempt.
However, the probability of picking up the second chip - be it blue or green - cannot be determined
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16 Apr 2013, 10:17
rajcools wrote:
C

1)Ratios are never really gud enough to find probability. Check with some values that are in ratio 3:4 . A is out
2) In this we have B and G = B+5 we can deduce the probability in terms of B. But we dont know the value of B

use B/B+5 = 3/4

and u r done

Your first statement " Ratios are never really gud enough to find probability" is not completely accurate. Ratios are sufficient, PROVIDED we are asked to find out the
probability of picking up either a green or a blue chip in the FIRST attempt.

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16 Apr 2013, 22:04
lot's of great algebra here, but if the hangup for people is wrapping your head around the insufficiency of statement one, I think picking test numbers is the easiest way. Any two sets of test numbers will bear out the insufficiency of this statement

that's how I did it anyway. Surely there's some other people out there like me whose vision starts to blur when they see 8 lines of algebraic expressions. It's probably better to be able to be comfortable with solving this algebraically, but get by however you can I say

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17 Apr 2013, 05:41
The bowl contains green and blue chips. What is the probability of drawing a blue chip in two successive trials if the chip drawn in the first trial is not returned to the bowl before the second trial?

1. The ratio of blue chips to green chips is 3:4
2. There are 5 more green chips than blue chips

For Case 1:
B:G = 3:4
So B = 3 and G = 4 => 2 successive = 3/7 * 2/6.....1
OR B = 6 and G = 8 => 2 successive = 6/14 * 4 *13....2

1 and 2 are not equal Hence Not Sufficient

For Case 2:
B = 2 and G = 7 => 2 successive = 2/9 * 1/8....1
OR B = 3 and G = 8 => 2 successive = 3/11 * 2/10....2

1 and 2 are not equal Hence Not Sufficient

Combining Case 1 and 2:
B = 15 and G = 20
we get a fixed answer for this condition

Hence C

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Re: M09 Q11   [#permalink] 17 Apr 2013, 05:41
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# M09 Q11

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