Find the probability of getting a Friday in a leap year.
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Now 364 is divisible by 7 and therefore there will be two excess week days in aleap year. The two excess week days can be (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday,Saturday), (Saturday, Sunday). So, the sample space S has 7 pairs of excess week days.
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A leap year has 366 days. If we take the weeks, it has 52. Altogether, 52 weeks and 2 days. So, in all possibility, we are going to have 52 Fridays. The case left is of the remaining 2 days.
Now, the two days in order could be
Thursday & FridayFriday & Saturday
There are in total 7 possibilities of these two days as the 1st day could be any one of the seven days and consequently the 2nd would be the adjacent day.
So, the probability of 53 Fridays turns out to be 2/7
What is the probability of 53 Fridays in a leap year?
A leap year has 366 days. If we take the weeks, it has 52. Altogether, 52 weeks and 2 days. So, in all possibility, we are going to have 52 Fridays. The case left is of the remaining 2 days.
Now, the two days in order could be
Thursday & FridayFriday & Saturday
There are in total 7 possibilities of these two days as the 1st day could be any one of the seven days and consequently the 2nd would be the adjacent day.
So, the probability of 53 Fridays turns out to be 2/7
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Prajjwal Pathak, BSc Computer Science & Physics, Banaras Hindu University (2021)
Answered Jan 14, 2018
1 year = 365 days
A leap year has 366 days
A year has 52 weeks.
Hence there will be 52 Sundays for sure.
52 weeks = 364 days
366 – 364 =2 days
In a leap year there will be 52 Sundays and 2 days will be left.
These 2 days can be:
Sunday, Monday
Monday, Tuesday
Tuesday, Wednesday
Wednesday, Thursday
Thursday, Friday
Friday, Saturday
Saturday, Sunday
Of these total 7 outcomes, the favourable outcomes are 2.
Hence the probability of getting 53 days 2/7.
—Thanking you
( Prajjwal pathak )
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Soumil Jain, former Associate at Indus Valley Partners (2017)
Answered Mar 23, 2018
There are 366 days in a leap year that is 2 more than 52 full weeks
52*7 + 2 = 366
Therefore now for the last 2 days, there could be 7 possibilities all equally likely. These are:
Monday, Tuesday
Tuesday, Wednesday
Wednesday, Thursday
Thursday, Friday
Friday, Saturday
Saturday, Sunday
Sunday, Monday
There are 2 cases out of the 7 cases above such that there is an extra friday.
Therefore the required probability here is 2 / 7
Now, the two days in order could be
Thursday & FridayFriday & Saturday
There are in total 7 possibilities of these two days as the 1st day could be any one of the seven days and consequently the 2nd would be the adjacent day.
So, the probability of 53 Fridays turns out to be 2/7
What is the probability of 53 Fridays in a leap year?
A leap year has 366 days. If we take the weeks, it has 52. Altogether, 52 weeks and 2 days. So, in all possibility, we are going to have 52 Fridays. The case left is of the remaining 2 days.
Now, the two days in order could be
Thursday & FridayFriday & Saturday
There are in total 7 possibilities of these two days as the 1st day could be any one of the seven days and consequently the 2nd would be the adjacent day.
So, the probability of 53 Fridays turns out to be 2/7
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Prajjwal Pathak, BSc Computer Science & Physics, Banaras Hindu University (2021)
Answered Jan 14, 2018
1 year = 365 days
A leap year has 366 days
A year has 52 weeks.
Hence there will be 52 Sundays for sure.
52 weeks = 364 days
366 – 364 =2 days
In a leap year there will be 52 Sundays and 2 days will be left.
These 2 days can be:
Sunday, Monday
Monday, Tuesday
Tuesday, Wednesday
Wednesday, Thursday
Thursday, Friday
Friday, Saturday
Saturday, Sunday
Of these total 7 outcomes, the favourable outcomes are 2.
Hence the probability of getting 53 days 2/7.
—Thanking you
( Prajjwal pathak )
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Soumil Jain, former Associate at Indus Valley Partners (2017)
Answered Mar 23, 2018
There are 366 days in a leap year that is 2 more than 52 full weeks
52*7 + 2 = 366
Therefore now for the last 2 days, there could be 7 possibilities all equally likely. These are:
Monday, Tuesday
Tuesday, Wednesday
Wednesday, Thursday
Thursday, Friday
Friday, Saturday
Saturday, Sunday
Sunday, Monday
There are 2 cases out of the 7 cases above such that there is an extra friday.
Therefore the required probability here is 2 / 7
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