A) 6, 3
B) 3, 6
C) 2, 7
D) None of these
Correct Answer: A
Solution :
[a] Let the number of red and blue balls be r and b, respectively. Then, the probability of drawing two red balls \[={{p}_{1}}=\frac{^{r}{{C}_{2}}}{^{r+b}{{C}_{2}}}=\frac{r(r-1)}{(r+b)(r+b-1)}\] The probability of drawing two blue balls is \[{{p}_{2}}=\frac{^{b}{{C}_{2}}}{^{r+b}{{C}_{2}}}=\frac{b(b-1)}{(r+b)(r+b-1)}\] The probability of drawing one red ad one blue ball \[={{p}_{3}}=\frac{^{r}{{C}_{1}}^{b}{{C}_{1}}}{^{r+b}{{C}_{2}}}=\frac{2br}{(r+b)(r+b-1)}\] By hypothesis, \[{{p}_{1}}=5{{p}_{2}}\] and \[{{p}_{3}}=6{{p}_{2}}.\]Therefore, \[r(r-1)=5b(b-1)\] and \[2br=6b(b-1)\] \[\Rightarrow r=6,b=3.\]
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