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Re: The bowl contains green and blue chips. What is the [#permalink]
1)
(3*m)/(4*m)=3/4
(3*m-1*3/4)/(4*m-1*4/3)
i don't know m, therefore insufficient

2) ratio of chips unknown

1&2)
4*m-3*m=5 ->
m=5
I can solve (3*m-1*3/4)/(4*m-1*4/3) for some numeric value
sufficient

C
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Re: The bowl contains green and blue chips. What is the [#permalink]
sondenso 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

I dont agree with the reasoning in OE!


I choose C.

1. The ratio is given, but without the # of total chips, we don't the probability. There could be 6 blue chips and 8 green chips, or 12 blue chips and 16 green chips.

Insufficient, rule out A & D.

2. Five more green chips to blue chips means there could be 1 blue chip, 6 green chips or 10 blue chips and 15 green chips.

Insufficient. Rule out B.

1 & 2 together: If there are 5 more green chips to blue chips, then we can find the total. Since the ratio of 3:4 is a difference of 1, we can multiply it by 5, resulting in 15:20, thus there are 5 more green chips than blue chips.

You can then solve for various probabilities.

C.
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Re: The bowl contains green and blue chips. What is the [#permalink]
Maple wrote:
1)
(3*m)/(4*m)=3/4
(3*m-1*3/4)/(4*m-1*4/3)


I do not understand your use of m-1 in your equation(s) above.
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Re: The bowl contains green and blue chips. What is the [#permalink]
RyanDe680 wrote:
I do not understand your use of m-1 in your equation(s) above.


1 chip times the probability that that chip will be removed.
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Re: The bowl contains green and blue chips. What is the [#permalink]
Maple wrote:
RyanDe680 wrote:
I do not understand your use of m-1 in your equation(s) above.


1 chip times the probability that that chip will be removed.


Excellent, thanks.
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Re: The bowl contains green and blue chips. What is the [#permalink]
RyanDe680 wrote:

Excellent, thanks.


you missed my error though. both should be divided by 7. :oops:
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Re: The bowl contains green and blue chips. What is the [#permalink]
Maple wrote:
RyanDe680 wrote:

Excellent, thanks.


you missed my error though. both should be divided by 7. :oops:


Yeah now you've lost me even more :oops:
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Re: The bowl contains green and blue chips. What is the [#permalink]
Maple wrote:
RyanDe680 wrote:

Yeah now you've lost me even more :oops:


Maple wrote:
1)
(3*m)/(4*m)=3/4
(3*m-1*3/7)/(4*m-1*4/7)
i don't know m, therefore insufficient

2) ratio of chips unknown

1&2)
4*m-3*m=5 ->
m=5
I can solve (3*m-1*3/7)/(4*m-1*4/7) for some numeric value
sufficient

C


The reason OE says A is not sufficient is "Although we know the probability that the first chip will be blue, we cannot compute the probability that the second chip will be blue."

If so, I think E is correct, not C
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Re: The bowl contains green and blue chips. What is the [#permalink]
sondenso wrote:

The reason OE says A is not sufficient is "Although we know the probability that the first chip will be blue, we cannot compute the probability that the second chip will be blue."

If so, I think E is correct, not C


OE is absolutely right. The probability is not computable. I just set up the formula to calculate the probability(actually the ratio between blue and green chips, which is similar). The formula is unsolvable without additional information(specifically the dummy variable m); hence the reason I deemed 1 insufficient.
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Re: The bowl contains green and blue chips. What is the [#permalink]
Do we have a conclusion what the OA is?

I have read all the comments and explanations esp on Statement (1). I am stilll wondering why Statement (1) was ruled out as insufficient.

a. Do we need to know the no. of green and blue chips in order to answer this question? My answer is NO. We just need to know the ratios. As given, the ratio of blue chips to green chips is 3:4. This means 3/7th is blue and 4/7th is green. Recall that the way ratio is written 3:4 means 3 out of 7 (3 blue and 4 green) and 4 out of 7 (3 blue and 4 green). After all, probability is ratio; the ratio of winning outcomes to total outcomes.

b. I saw some explanations quoted eg. "blue / greeen balls = 3 / 4 (ratio of balls)" and "(3*m)/(4*m)=3/4". Ratio 3:4 can be shown this way

\(\frac{3}{7}\) divided by \(\frac{4}{7}\)
= \(\frac{3}{7}\) x \(\frac{7}{4}\)
= 3 : 4

Anyone else saw the above as I see it?
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Re: The bowl contains green and blue chips. What is the [#permalink]
jimmylow wrote:
Do we have a conclusion what the OA is?

I have read all the comments and explanations esp on Statement (1). I am stilll wondering why Statement (1) was ruled out as insufficient.

a. Do we need to know the no. of green and blue chips in order to answer this question? My answer is NO. We just need to know the ratios. As given, the ratio of blue chips to green chips is 3:4. This means 3/7th is blue and 4/7th is green. Recall that the way ratio is written 3:4 means 3 out of 7 (3 blue and 4 green) and 4 out of 7 (3 blue and 4 green). After all, probability is ratio; the ratio of winning outcomes to total outcomes.

b. I saw some explanations quoted eg. "blue / greeen balls = 3 / 4 (ratio of balls)" and "(3*m)/(4*m)=3/4". Ratio 3:4 can be shown this way

\(\frac{3}{7}\) divided by \(\frac{4}{7}\)
= \(\frac{3}{7}\) x \(\frac{7}{4}\)
= 3 : 4

Anyone else saw the above as I see it?



299/700 is very different from 2/7.



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