Question

Q.5 what is the probability that a leap year selected at random will contain either 53 thursdays or 53 fridays?

Answer 1

There are 7 possible leap years:

Leap year starts on a Monday, result: 53 Mondays and 53 Tuesdays.

Leap year starts on Tuesday: 53 Tuesdays and 53 Wednesdays.

Leap yr. begins Wednesday: 53 Wednesdays and 53 Thursdays.

Leap yr. begins Thurs.: 53 Thursdays and 53 Fridays

Leap year begins Fri.: 53 Fridays and 53 Saturdays

Leap year begins Sat.: 53 Saturdays and 53 Sundays

Leap year begins Sun.: 53 Sundays and 53 Mondays.

Answer 2

There are 366 days in a leap year, of which 364 days comprise of 52 weeks.

Now, we are left with 2 excess days in a leap year. These 2 days can be as follows:

(Sunday, Monday)

(Monday, Tuesday)

(Tuesday, Wednesday)

(Wednesday, Thursday)

(Thursday, Friday)

(Friday, Saturday)

(Saturday, Sunday)

Therefore, the sample space 'S' will be these 7 pairs of excess days.

Therefore, n(S) = 7;

Now, what we need is the event E to be either 53 thursdays or 53 fridays. There are 3 such cases in sample space S, where we have the 53rd thursday or friday.

Therefore, n(E) = 3;

Hence, the probability of the desired event will be : n(E)/n(S);

=3/7