A) \[9:6:4\]
B) \[8:6:5\]
C) \[10:5:4\]
D) None of these
Correct Answer: A
Solution :
| [a] Probability of getting a composite number is \[2/6=1/3\] |
| Probability that A will win the game is |
| \[\left( \frac{2}{3} \right)\left( \frac{1}{3} \right)+\left( \frac{2}{3} \right)\left( \frac{2}{3} \right)\left( \frac{2}{3} \right)\left( \frac{2}{3} \right)\left( \frac{1}{3} \right)\]\[+\left( \frac{2}{3} \right)\left( \frac{2}{3} \right)\left( \frac{2}{3} \right)\left( \frac{2}{3} \right)\left( \frac{2}{3} \right)\left( \frac{2}{3} \right)\left( \frac{2}{3} \right)\left( \frac{1}{3} \right)+....\] |
| \[=\frac{\frac{1}{3}}{1-\frac{8}{27}}=\left( \frac{1}{3}\times \frac{27}{19} \right)=\frac{9}{19}\] |
| Probability that B will win the game is |
| \[\left( \frac{2}{3} \right)\left( \frac{1}{3} \right)+\left( \frac{2}{3} \right)\left( \frac{2}{3} \right)\left( \frac{2}{3} \right)\left( \frac{2}{3} \right)\left( \frac{1}{3} \right)\] |
| \[+\left( \frac{2}{3} \right)\left( \frac{2}{3} \right)\left( \frac{2}{3} \right)\left( \frac{2}{3} \right)\left( \frac{2}{3} \right)\left( \frac{2}{3} \right)\left( \frac{2}{3} \right)\left( \frac{1}{3} \right)+...\] |
| \[=\frac{\frac{2}{9}}{1-\frac{8}{27}}=\left( \frac{2}{9}\times \frac{27}{19} \right)=\frac{6}{19}\] |
| Probability that C will win the game is |
| \[\left( \frac{2}{3} \right)\left( \frac{2}{3} \right)\left( \frac{1}{3} \right)+\left( \frac{2}{3} \right)\left( \frac{2}{3} \right)\left( \frac{2}{3} \right)\left( \frac{2}{3} \right)\left( \frac{1}{3} \right)\] |
| \[+\left( \frac{2}{3} \right)\left( \frac{2}{3} \right)\left( \frac{2}{3} \right)\left( \frac{2}{3} \right)\left( \frac{2}{3} \right)\left( \frac{2}{3} \right)\left( \frac{2}{3} \right)\left( \frac{2}{3} \right)\left( \frac{1}{3} \right)+...\] |
| \[=\frac{\frac{4}{27}}{1-\frac{8}{27}}=\left( \frac{4}{27}\times \frac{27}{19} \right)=\frac{4}{19}\] |
| So required ratio is \[9:6:4,\] |
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