What is the probability that a leap year, selected at random will contain (i) 53 Fridays (ii) 53 Mondays and 53 Tuesdays (iii) 53 Sundays and 53 Thursdays. [4 MARKS]
What is the probability that a leap year, selected at random will contain (i) 53 Fridays (ii) 53 Mondays and 53 Tuesdays (iii) 53 Sundays and 53 Thursdays. [4 MARKS]
Sample space: 1 Mark
Each subpart: 1 Mark each
A leap year has 366 days, which works out to 52 full weeks and two additional days.
These two additional days could be:
Sunday - Monday
Monday - Tuesday
Tuesday - Wednesday
Wednesday - Thursday
Thursday - Friday
Friday - Saturday
Saturday - Sunday.
These seven outcomes are equally likely.
(i) Thus, the probability that one of these two days is Friday is 27.
(ii) There’s only one outcome above which accounts for 53 Mondays and 53 Tuesdays. Therefore, the probability is 17.
(iii) There is no outcome above which will account for 53 Sundays and 53 Thursdays. So the probability is 0.
Sample space: 1 Mark
Each subpart: 1 Mark each
A leap year has 366 days, which works out to 52 full weeks and two additional days.
These two additional days could be:
Sunday - Monday
Monday - Tuesday
Tuesday - Wednesday
Wednesday - Thursday
Thursday - Friday
Friday - Saturday
Saturday - Sunday.
These seven outcomes are equally likely.
(i) Thus, the probability that one of these two days is Friday is 27.
(ii) There’s only one outcome above which accounts for 53 Mondays and 53 Tuesdays. Therefore, the probability is 17.
(iii) There is no outcome above which will account for 53 Sundays and 53 Thursdays. So the probability is 0.
