A) \[\frac{1}{4}\]
B) \[\frac{2}{7}\]
C) \[\frac{1}{3}\]
D) \[\frac{3}{8}\]
Correct Answer: C
Solution :
| Suppose the green ball goes to bin i, for some \[i\ge 1.\] |
| The probability of this occurring is \[\frac{1}{{{2}^{i}}}\] |
| Given that this occurs, the probability that the red ball goes in a higher numbered bin is |
| \[\frac{1}{{{2}^{i+1}}}+\frac{1}{{{2}^{i+2}}}+...=\frac{1}{{{2}^{i}}}\] |
| Probability that the green ball goes to bin i and the red ball goes in a bin greater than i is \[{{\left( \frac{1}{2} \right)}^{2}}-\frac{1}{{{2}^{2}}}=\frac{1}{{{4}^{i}}}\] |
| Required probability \[\sum\limits_{i=1}^{\infty }{\frac{1}{{{4}^{i}}}}\]\[=\frac{1}{4}+\frac{1}{{{4}^{2}}}+\frac{1}{{{4}^{3}}}...=\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{1}{3}\] |
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