What is the probability that a non-leap year has 53 Sundays?

A  \(\frac{6}7\)

B  \(\frac{1}7\)

C  \(\frac{5}7\)

D  None of these


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Total numbers of elementary events are: 7 

Let E be the event of having 53 Sundays 

Note- a non-leap year has 365 days or 52 weeks and 1 day. This one day could be any day amongst Sunday, Monday, Tuesday, Wednesday, Thursday, Friday or Saturday. Out of these favorable event is one; Sunday. 

Number of favorable outcome is = 1 

P (Sunday) = P (E) = \(\frac{1}7\)

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