A) \[\frac{3}{28}\]
B) \[\frac{2}{28}\]
C) \[\frac{7}{28}\]
D) \[\frac{5}{28}\]
Correct Answer: D
Solution :
We know a leap year is fallen within 4 years, so its probability is \[\frac{25}{100}=\frac{1}{4}\] 53rd Sunday is a leap year \[=\frac{1}{4}\times \frac{2}{7}\,=\,\frac{2}{28}\] Similarly probability of 53rd Sunday in a non-leap year \[=\frac{75}{100}\times \frac{1}{7}\,=\frac{3}{4}\times \frac{1}{7}=\frac{3}{28}\] \ Required probability \[=\frac{2}{28}+\frac{3}{28}=\frac{5}{28}.\]
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