what is the probability that a leap year selected at random will contain either 53 Thursday or 53 Friday?
Answers
We know that there are 366 days in a leap year.
= > 52 weeks + 2 days.
These 2 days can b:
(a) Monday, Tuesday
(b) Tuesday, Wednesday,
(c) Wednesday, Thursday
(d) Thursday, Friday
(e) Friday, Saturday
(f) Saturday, Sunday
(g) Sunday, Monday.
= > Let A be the event that a leap year contains 53 Thursdays.
n(A) = {Wednesday, Thursday},{Thursday, Friday}}.
= 2
Required probability p(A) = n(A)/n(S)
= 2/7.
= > Let B be the event that a leap year contains 53 Fridays.
n(B) = {Thursday, Friday},{Friday, Saturday}
= 2.
Required probability p(B) = n(B)/n(S)
= 2/7.
= > Let C be the event that where Thursday and Friday occur simultaneously.
n(C) = {Thursday,Friday}.
Required probability P(C) = n(C)/n(S)
= 1/7.
Therefore, the required probability = P(A) + P(B) - P(C)
= (2/7) + (2/7) - (1/7)
= 3/7.
Hope it helps!
The answer is 3/7
GIVEN
Leap year having either 53 Thursdays or 53 Fridays.
TO FIND
The probability.
SOLUTION
We can simply solve the above problem as follows;
We know that,
Number of days in leap year = 356 days
356 days = 52 weeks and 2 days.
The two days can be of following combinations;
(Monday, Tuesday),(Tuesday, Wednesday),(Wednesday, Thursday),(Thursday, Friday), (Friday, Saturday),(Saturday, Sunday), ( Sunday, Monday)
Number of combinations = 7
Probability that the year has 53 Thursdays = Combinations having Thursday/Total combinations
= 2/7
Probability that the year has 53 Friday's = Combinations having Friday/Total combination
= 2/7
Probability that the combination has both Thursday and Friday = 1/7
Required probability = 2/7 + 2/7 - 1/7
= 3/7
Hence, The answer is 3/7
#SPJ2