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14.Probability
hard
The probability that in a year of the $22^{nd}$ century chosen at random there will be $53$ Sundays is
A
$\frac{3}{{28}}$
B
$\frac{2}{{28}}$
C
$\frac{7}{{28}}$
D
$\frac{5}{{28}}$
Solution
(d) 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 $53^{rd}$ 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}}.$
Standard 11
Mathematics
