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The bowl contains green and blue chips. What is the [#permalink]
10 May 2008, 01:15
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
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! _________________
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!
IMO C should be the answer. Here is the explanation..
Statement 1 does not tells us what is the total number of balls.. therefore u cant take out the probability
Statement 2 alone does not tell u what is the number of bule chips and accordingly u cant calculate the total number of balls and hence the probability of blue balls
Statement 1 and 2 put together .....
u can calculate the number of balls ..here is how
blue / greeen balls = 3 / 4 (ratio of balls)
blue chips = x and green chips = x 5
accordingly solving the above mentioned two equations u will get x = 15 and total number of balls as 35 and accordingly u can calculate the required probability.
Hope this helps .
what is the OA and the OE to the problem _________________
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!
My answer is A.
Statement (1) alone tell us the ratio of blue to green is 3:4. Blue = 3/7 and Green = 4/7. Thus, probability of (blue, blue) = \(\frac{3}{7}\) * \(\frac{2}{6}\) = \(\frac{1}{7}\). Sufficient. Eliminate B, C, E.
Statement (2) alone tells us that there are 5 more green chips than blue but does not tell us how many are blue chips. Neither does the question stem contains any clue. Insufficient. Eliminate D. _________________
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.
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 _________________
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.
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? _________________
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