What is the probability of getting 53 Wednesday in leap year
Answers
Our sample space is S : {Monday-Tuesday, Tuesday-Wednesday, Wednesday-Thursday,..., Sunday-Monday}
Number of elements in S = n(S) = 7
What we want is a set A (say) that comprises of the elements Saturday-Sunday and Sunday-Monday i.e. A : {Saturday-Sunday, Sunday-Monday}
Number of elements in set A = n(A) = 2
By definition, probability of occurrence of A = n(A)/n(S) = 2/7
Answer:
2/7
Step-by-step explanation:
1 year = 365 days
A leap year has 366 days
A year has 52 weeks. Hence there will be 52 Wednesdays.
52 weeks = 52*7 = 364 days
366-364 = 2 days
In a leap year there will be 52 Wednesdays 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 the total 7 outcomes the favorable outcomes are 2.
Hence the probability of getting 53 Wednesdays in a leap year = 2/7
Hope it helpzzz