Economist GMAT
There were \(r\) red balls and \(y\) yellow balls in a bag. Three red balls and four yellow ones were added to the bag. What is the probability that a yellow ball and then a red ball will be selected if Jerry pulls out 2 balls at random, and puts the first ball back before pulling the second ball?
A. \(\frac{y+4}{y+r+7} \cdot \frac{r+3}{y+r+7}\)
B. \(\frac{y+3}{y+r+7} + \frac{r+4}{y+r+6}\)
C. \(\frac{y+3}{y+r+7} \cdot \frac{r+4}{y+r+6}\)
D. \(\frac{y+3}{y+r+7} \cdot \frac{r+3}{y+r+7}\)
E. \(\frac{y+3}{y+r+7} + \frac{r+3}{y+r+6}\)
OA A
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There were \(r\) red balls and \(y\) yellow balls in a bag.
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GMATGuruNY
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Let r=0 and y=0, implying that the bag is EMPTY before the addition of 3 red balls and 4 yellow balls.AAPL wrote:Economist GMAT
There were \(r\) red balls and \(y\) yellow balls in a bag. Three red balls and four yellow ones were added to the bag. What is the probability that a yellow ball and then a red ball will be selected if Jerry pulls out 2 balls at random, and puts the first ball back before pulling the second ball?
A. \(\frac{y+4}{y+r+7} \cdot \frac{r+3}{y+r+7}\)
B. \(\frac{y+3}{y+r+7} + \frac{r+4}{y+r+6}\)
C. \(\frac{y+3}{y+r+7} \cdot \frac{r+4}{y+r+6}\)
D. \(\frac{y+3}{y+r+7} \cdot \frac{r+3}{y+r+7}\)
E. \(\frac{y+3}{y+r+7} + \frac{r+3}{y+r+6}\)
P(yellow ball) = 4/7. (Of the 7 balls, 4 are yellow.)
P(red ball) = 3/7. (Of the 7 balls, 3 are red.)
To combine these probabilities, we multiply:
4/7 * 3/7
The correct answer must yield the product above when r=0 and y=0.
Only A works:
\(\frac{y+4}{y+r+7} \cdot \frac{r+3}{y+r+7}\) = \(\frac{0+4}{0+0+7} \cdot \frac{0+3}{0+0+7}\) = \(\frac{4}{7} \cdot \frac{3}{7}\)
The correct answer is A.
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We have y+4 yellow balls and r+3 red balls, and r+y+7 balls in total. So the probability of picking, with replacement, first a yellow then a red ball will be (y+4)/(r+y+7) * (r+3)/(r+y+7), which is answer A.
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The probability that the first ball is yellow is (y + 4)/(y + r + 7). The probability that the second ball is red, after the first ball is put back into the bag, is (r + 3)/(y + r + 7). Therefore, the overall probability is:AAPL wrote:Economist GMAT
There were \(r\) red balls and \(y\) yellow balls in a bag. Three red balls and four yellow ones were added to the bag. What is the probability that a yellow ball and then a red ball will be selected if Jerry pulls out 2 balls at random, and puts the first ball back before pulling the second ball?
A. \(\frac{y+4}{y+r+7} \cdot \frac{r+3}{y+r+7}\)
B. \(\frac{y+3}{y+r+7} + \frac{r+4}{y+r+6}\)
C. \(\frac{y+3}{y+r+7} \cdot \frac{r+4}{y+r+6}\)
D. \(\frac{y+3}{y+r+7} \cdot \frac{r+3}{y+r+7}\)
E. \(\frac{y+3}{y+r+7} + \frac{r+3}{y+r+6}\)
OA A
(y + 4)/(y + r + 7) * (r + 3)/(y + r + 7)
Answer: A
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