A bag contains only red and white balls. The probability of drawing tw

Question

A bag contains only red and white balls. The probability of drawing two red balls consecutively is  \frac{2}{5} . However, when 6 red balls are replaced with 6 white balls, the probability of drawing two consecutive red balls decreases to  \frac{3}{20} .

Select the number of red balls in the bag prior to the replacement under "Red" and the number of white balls in the bag after the replacement under "White". Make only two selections, one from each column.

chetan2u · Accepted Answer

Let there be r red balls and total t balls.

The probability of drawing two red balls consecutively is  \frac{2}{5} . 
Thus P = rC2/nC2 = 2/5 ...... r(r-1)/n(n-1) =2/5 ......(i)

However, when 6 red balls are replaced with 6 white balls, the probability of drawing two consecutive red balls decreases to  \frac{3}{20} .
Red balls remaining = r-6, while total remain n
Thus P= (r-6)C2/nC2 = (r-6)(r-7)/n(n-1) = 3/20......(ii)

Divide i by ii
r(r-1)/(r-6)(r-7) = 40/15 = 8/3

Substitute the options to see what fits in for r. 
r as 16 gives 16*15/10*9 or 8/3

Next substitute r in (i) to get n
16*15/n(n-1) = 2/5
n(n-1) = 8*15*5 = 8*5*3*5 = 5*5*8*3 = 25*24
Hence total =25 and white = 25-16 = 9
After replacing Six red with white, white =9+6 = 15

dsahay · Answer

Please give the solution

Bismuth83 · Answer

1. We are asked to  find the number of red balls in the bag prior to the replacement and the number of white balls in the bag after the replacement .

2. Let r be the number of red balls and n be the total number of balls. The probability of picking two red balls consecutively is equal to  \frac{number \ of \ ways \ to \ pick \ 2 \ red \ balls}{number \ of \ ways \ to \ pick \ 2 \ balls} .

3. It is known that this probability prior to the replacement is equal to  \frac{2}{5} . So,   \frac{2}{5} = \frac{number \ of \ ways \ to \ pick \ 2 \ red \ balls}{ number \ of \ ways \ to \ pick \ 2 \ balls } = \frac{r \choose 2}{n \choose 2} .

4. After the replacement, the number of red balls becomes r-6. Since the new probability is  \frac{3}{20} , we have: \frac{3}{20} = \frac{number \ of \ ways \ to \ pick \ 2 \ red \ balls}{number \ of \ ways \ to \ pick \ 2 \ balls} = \frac{r-6 \choose 2}{n \choose 2}  .

5. Our question is asking for r and n-(r-6). If we can find r, then n-(r-6) can also be found too.

6. To do this, we can divide the two previous equations:  \frac{\frac{2}{5}}{\frac{3}{20}} = \frac{\frac{r \choose 2}{n \choose 2}}{\frac{r-6 \choose 2}{n \choose 2}} \rightarrow \frac{8}{3} = \frac{r \choose 2}{r-6 \choose 2} \rightarrow 3r(r-1) = 8(r-6)(r-7) \rightarrow 3r^2 - 3r = 8r^2 - 104r + 336 \rightarrow 5r^2 - 101r + 336 = 0 \rightarrow (5r - 21)(r - 16) = 0  .

7. Then,  r = 16 \ or \ 21/5 , however, r must be an integer so it’s equal to 16. Now let’s find n:  \frac{3}{20} = \frac{10C2}{nC2} \rightarrow n(n-1) = 600 \rightarrow n = 25 . So, the number of white balls after the replacement is equal to n-(r-6) = 25-(16-6) = 15.

8. Our answer will be   16 red balls before the replacement   and   15 white balls after the replacement  .

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=83049 Harsha GMAT-Club-Forum-kb2gfp76.png

kayarat600 · Answer

This seems like a lengthy question not sure if it's feasible under the time seems like a leave for later question unless you can plug options effectively.

Bismuth83 · Answer

I agree with you that the question is lengthy, however I think it's feasible under the time. Most of the problem is representing through algebra and simplifying which is straightforward but just requires a "quick hand." The only step that I think could take some time is figuring out that you have to divided to (efficiently) solve the system of equations.

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A bag contains only red and white balls. The probability of drawing tw

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GMAT Two-part Analysis (TPA) Questions

A bag contains only red and white balls. The probability of drawing tw

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