Question

The probability that a leap year will have $53$ fridays or $53$ saturdays, is:

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

Answer 1

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

In a leap year there are 366 days, in which 52 weeks and two days. The combination if two days may be
Sun-Mon
Mon-Tue
Tue-Wed
Wed-Thur
Thur-Fri
Fri-Sat
Sat-Sun
$P\left(53Fri\right)=\frac{2}{7}$, $P\left(53sat\right)=\frac{2}{7}$
One combination is common ie, $(Fri-Sat)$
$\therefore$ $P\left(53Friand53Sat\right)=\frac{1}{7}$
$\therefore$ $P\left(53Friand53Sat\right)=P\left(53Fri\right)+P\left(53Sat\right)-P\left(53Friand53Sat\right)$
$=\frac{2}{7}+\frac{2}{7}-\frac{1}{7}$
$=\frac{3}{7}$