Question

find the probability that a non leap year selected at random will contain only 52 Sundays

Answer 1

$\huge\boxed{ \fcolorbox{red}{aqua}{hiii \: mate}}$


$\huge \mathfrak{here \: is \: ur \: answer}$


We have ,

365 days = 52 weeks and 1 day

Thus, a non leap year has always 52 Sundays.

The remaining can be
✴Monday and Tuesday
✴Tuesday and Wednesday
✴Wednesday and Thursday
✴Thursday and Friday
✴Friday and Saturday
✴Saturday and Sunday
✴ Sunday and Monday

Clearly, there are seven elementary events associated with this random experiment.

Let A be the event that a non leap year has 52 Sundays.

Clearly, the event A will happen if the last one day of the non leap year is either Sunday and Monday.

Therefore, Favourable event = 1

Hence ,required probability = $\frac{1}{7}$


$\huge\bold{Answer = \frac{1}{7} }$