Question

In a non-leap year the probability of getting $53$ Sundays or $53$ Tuesdays or $53$ Thursdays is.

• A. $\frac{1}{7}$
• B. $\frac{2}{7}$
• C. $\frac{3}{7}$
• D. $\frac{4}{7}$

Answer 1

The correct option is C $\frac{3}{7}$

Non-leap year has $365$ days

$1$ year $=52$ weeks

$=52\times 7days$

$=364$ days [It contains $52$ Sundays, Mondays and so on].

$\therefore 1$ day will be left sample space for this

$1$ day $=\{$Mon, Tue, Wed, Thurs, Fri, Sat, Sun$\}$

$P[53$ Sundays$]=\frac{1}{7}$

$P[53$ Tuesdays$]=\frac{1}{7}$

$P[53$ Thursdays$]=\frac{1}{7}$

$P(53$ Sundays or $53$ Tuesdays or $53$ Thursdays$)$

$=P(53$ Sundays$)+P(53$ Tuesdays$)+P(53$ Thursdays$)$

$=\frac{1}{7}+\frac{1}{7}+\frac{1}{7}$

$=\frac{3}{7}$