A) 3 : 4
B) 4 : 3
C) 2 : 1
D) 1 : 2
Correct Answer: C
Solution :
[c] Probability of A winning [A can win in 1st or 3rd or 5th ?games if B loses 2nd or 4th or ?games] \[=\frac{p}{p+q}+{{\left( \frac{q}{p+q} \right)}^{2}}.\frac{p}{p+q}+{{\left( \frac{q}{p+q} \right)}^{4}}.\frac{p}{p+q}+...\] \[=\frac{\frac{p}{p+q}}{1-{{\left( \frac{q}{p+q} \right)}^{2}}}\left[ In\,\inf inite\,G.P.S=\frac{a}{1-r} \right]\] \[=\frac{p(p+q)}{{{(p+q)}^{2}}-{{q}^{2}}}\] Probability of B winning \[=1-\frac{p(p+q)}{{{(p+q)}^{2}}-{{q}^{2}}}=\frac{{{(p+q)}^{2}}-{{q}^{2}}-p(p+q)}{{{(p+q)}^{2}}-{{q}^{2}}}\] Given \[P(A)=3P(B)\] \[\Rightarrow p(p+q)=3[{{(p+q)}^{2}}-{{q}^{2}}-p(p+q)]\] \[\Rightarrow 4p(p+q)=3(p+2q).p\] \[\Rightarrow 4p+4q=3p+6q\Rightarrow p=2p\] \[\frac{p}{q}=2\] Or \[p:q=2:1\]
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