What is the probability that an ordinary year has 53 Sundays?

1.	What is the probability that an ordinary year has 53 Sundays?
Answer» Given: an ordinary year which includes 52 weeks and one day Formula: P(E) = \(\frac{favourable \ outcomes}{total \ possible \ outcomes}\) so, we have to determine the probability of that one day being Sunday Total number of possible outcomes are 7 Therefore n(S)=7 E= {M, T, W, T, F, S, SU} n(E) = 1 P(E) = \(\frac{n(E)}{n(S)}\) P(E) = \(\frac{1}{7}\)

Answer»

Given: an ordinary year which includes 52 weeks and one day

Formula: P(E) = \(\frac{favourable \ outcomes}{total \ possible \ outcomes}\)

so, we have to determine the probability of that one day being Sunday

Total number of possible outcomes are 7

Therefore n(S)=7

E= {M, T, W, T, F, S, SU}

n(E) = 1

P(E) = \(\frac{n(E)}{n(S)}\)

P(E) = \(\frac{1}{7}\)

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What is the probability that an ordinary year has 53 Sundays?

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