What is the probability of having 53 monday or 53 tuesday in the year 2020
Answers
ANSWER
A leap year has 366 days. Now 364 is divisible by 7 and therefore there will be two excess weekdays in a leap year.
The two excess weekdays 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 excessweekdays. i.e. n(S)=7
Now we want the desired event E to have 53 Mondays and 53 Tuesdays. E consists of only one pair in S which is (Monday, Tuesday). So, n(E)=1
Hence, the probability that a leap year will contain 53 Mondays and 53 Tuesdays =
n(S)
n(E)
=
7
1
Step-by-step explanation:
ANSWER
A leap year has 366 days. Now 364 is divisible by 7 and therefore there will be two excess weekdays in a leap year.
The two excess weekdays 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 weekdays. i.e. n(S)=7
Now we want the desired event E to have 53 Mondays and 53 Tuesdays. E consists of only one pair in S which is (Monday, Tuesday). So, n(E)=1
Hence, the probability that a leap year will contain 53 Mondays and 53 Tuesdays =
n(S)
n(E)
=
7
1
ANSWER
A leap year has 366 days. Now 364 is divisible by 7 and therefore there will be two excess weekdays in a leap year.
The two excess weekdays 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 weekdays. i.e. n(S)=7
Now we want the desired event E to have 53 Mondays and 53 Tuesdays. E consists of only one pair in S which is (Monday, Tuesday). So, n(E)=1
Hence, the probability that a leap year will contain 53 Mondays and 53 Tuesdays =
n(S)
n(E)
=
7
1