There were $r$ red balls and $y$ yellow balls in a bag.

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AAPL: Moderator; Posts: 2749; Joined: Sun Oct 29, 2017 2:08 pmFollowed by: 2 members

by AAPL » Sat Aug 24, 2019 10:05 am

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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

Source: Beat The GMAT — Problem Solving | Sat Aug 24, 2019 10:05 am

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sat Aug 24, 2019 12:12 pm

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. $\frac{y+3}{y+r+7} + \frac{r+3}{y+r+6}$

Let r=0 and y=0, implying that the bag is EMPTY before the addition of 3 red balls and 4 yellow balls.

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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by Ian Stewart » Sun Aug 25, 2019 8:18 am

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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by Scott@TargetTestPrep » Tue Aug 27, 2019 5:19 pm

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

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:

(y + 4)/(y + r + 7) * (r + 3)/(y + r + 7)

Answer: A

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