The probability of getting 53 Fridays in a Leap year isA) 1/7B) 2/7C)

1.	The probability of getting 53 Fridays in a Leap year isA) 1/7B) 2/7C) 3/7D) 4/7
Answer» Correct option is: B) \(\frac 27\) Number of days in a leap year = 366. \(\because\) \(\frac {366}7 = 52 + \frac 27 = \) 52 weeks and 2 days remaining. Hence, in a leap year there is definitely 52 Fridays but 53 Fridays depends on 2 remaining days. There are total 7 outcomes for these two remaining days. They can be (i) Sunday & Monday, (ii) Monday & Tuesday (iii) Tuesday & Wednesday (iv) Wednesday & Thursday (v) Thursday & Friday (vi) Friday & Saturday (vii) Saturday & Sunday But favourable outcomes for 53 Fridays are Thursday & Friday or Friday & Saturday. Hence, there are 2 favourable outcomes for 53 Fridays out of 7 total outcomes. \(\therefore\) Probability of getting 53 Fridays in a leap year is P = \(\frac {Total \, favourable \, outcome}{Total\, possible \, outcomes}\) = \(\frac 27\) Correct option is: B) \(\frac{2}{7}\)

The probability of getting 53 Fridays in a Leap year isA) 1/7B) 2/7C) 3/7D) 4/7

Answer»

Correct option is: B) \(\frac 27\)

Number of days in a leap year = 366.

\(\because\) \(\frac {366}7 = 52 + \frac 27 = \) 52 weeks and 2 days remaining.

Hence, in a leap year there is definitely 52 Fridays but 53 Fridays depends on 2 remaining days.

There are total 7 outcomes for these two remaining days.

They can be (i) Sunday & Monday, (ii) Monday & Tuesday (iii) Tuesday & Wednesday

(iv) Wednesday & Thursday (v) Thursday & Friday (vi) Friday & Saturday

(vii) Saturday & Sunday

But favourable outcomes for 53 Fridays are Thursday & Friday or Friday & Saturday.
Hence, there are 2 favourable outcomes for 53 Fridays out of 7 total outcomes.

\(\therefore\) Probability of getting 53 Fridays in a leap year is

P = \(\frac {Total \, favourable \, outcome}{Total\, possible \, outcomes}\) = \(\frac 27\)

Correct option is: B) \(\frac{2}{7}\)

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