A) 1 : 1
B) 1 : 2
C) 2 : 1
D) None of these
Correct Answer: C
Solution :
Let \[W\] denote the event of drawing a white ball at any draw and \[B\] that for a black ball. Then \[P(W)=\frac{a}{a+b},\] \[P(B)=\frac{b}{a+b}\] \[P(A\] wins the game)\[=P(W\]or\[BBW\]or\[BBBBW\]or ?..) \[=P(W)+P(B)P(B)P(W)+P(B)P(B)P(B)P(B)P(W)+\,......\] \[=\frac{P(W)}{1-P{{(B)}^{2}}}=\frac{\frac{a}{a+b}}{1-\frac{{{b}^{2}}}{{{(a+b)}^{2}}}}=\frac{a(a+b)}{{{a}^{2}}+2ab}=\frac{(a+b)}{a+2b}\] Also \[P(B\] wins the game)\[=1-\frac{a+b}{a+2b}=\frac{b}{a+2b}\] According to the given condition, \[\frac{a+b}{a+2b}=3.\frac{b}{a+2b}\Rightarrow a=2b\Rightarrow a:b=2:1.\]You need to login to perform this action.
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