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Bismuth83
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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.
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Harsha
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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.
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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.
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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