What is the probability that a leap year selected at random will contain either 53 thursdays or 53 Friday's?
What is the probability that a leap year selected at random will contain either 53 thursdays or 53 Friday's?
A leap year has 366 days. Now 364 is divisible by 7 and therefore there will be two excess week days Ina leap 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. i.e. n(S) = 7.
Now we want the desired event E to be 53 Thursdays or 53 Fridays. E consists of the pairs (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday). So, n(E) = 3.
Hence, the probability that a leap year selected at random will contain either 53 Thursdays or 53 Fridays = n(E)/n(S) = 3/7
A leap year has 366 days. Now 364 is divisible by 7 and therefore there will be two excess week days Ina leap 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. i.e. n(S) = 7.
Now we want the desired event E to be 53 Thursdays or 53 Fridays. E consists of the pairs (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday). So, n(E) = 3.
Hence, the probability that a leap year selected at random will contain either 53 Thursdays or 53 Fridays = n(E)/n(S) = 3/7
