MCQ
The probability that a non leap year selected at random will have $53$ Sundays is :
- A$\frac{4}{7}$
- B$\frac{3}{7}$
- ✓$\frac{1}{7}$
- D$\frac{2}{7}$
✓
Answer
Correct option: C.
$\frac{1}{7}$
Non $-$ leap year contains $365$ days $= 364$ days $+\ 1$ day $=\Big(\frac{364}{7}\Big)$ weeks $+\ 1$ day $= 52$ weeks $+\ 1$ remaining day $= 52$ Sundays $+\ 1$ remaining day
We will have $53$ Sundays if $1$ remaining day is a Sunday.
Possible outcomes $= \{\text{(Monday), (Tuesday), (Wednesday), (Thursday), (Friday), (Saturday), (Sunday)\}}$
Number of Total outcomes $= 7$
Number of possible outcomes $= 1$
$\therefore$ Required Probability $=\frac{\text{Possible outcomes}}{\text{Total outcomes}}=\frac{1}{7}$
We will have $53$ Sundays if $1$ remaining day is a Sunday.
Possible outcomes $= \{\text{(Monday), (Tuesday), (Wednesday), (Thursday), (Friday), (Saturday), (Sunday)\}}$
Number of Total outcomes $= 7$
Number of possible outcomes $= 1$
$\therefore$ Required Probability $=\frac{\text{Possible outcomes}}{\text{Total outcomes}}=\frac{1}{7}$
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