Question

If a leap year is selected at random, then the chance that it will contain 53 Tuesdays, is

• A. $\frac{1}{5}$
• B. $\frac{1}{7}$
• C. $\frac{2}{7}$
• D. $\frac{2}{5}$

Answer 1

Answer: (C)

In a leap year, there are 366 days having 52 weeks and 2 extra days.
In 52 weeks, there are 52 Tuesday.
Therefore, the probability that the leap year will contain 53 Tuesday is equal to the probability that from the remaining 2 days will one be Tuesday.
For the remaining 2 days there are following 7 cases
(i) Sunday and Monday
(ii) Monday and Tuesday
(iii) Tuesday and Wednesday
(iv) Wednesday and Thursday
(v) Thursday and Friday
(vi) Friday and Saturday
(vii) Saturday and Sunday
Out of these seven possibilities, only two i.e., (ii) and (iii) are favourable to the events ' 53 Tuesday'.
$\therefore$ Total number of cases $=7$ and favourable cases $=2$.
Here, required probability $=\frac{2}{7}$