Question

Find the probability that a leap year selected at random will contain 53 Sundays.

Answer 1

Answer:

1/365

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Answer 2

$\rm \huge{ \underline{ \underline{ \red{ \: \: answer \: \: }}}}$

We know that a normal year has 365 days with 7 days comprises a week.

Thus, Number of weeks in normal year =365÷7 = 52 weeks and 1 left a day(which is 52 Mondays, 52 Tuesdays, 52 Wednesdays, 52 Thursdays, 52 Fridays, 52 Saturdays, 52 Sundays)

A leap year has 366 days in which 2 days are left with 52 weeks.

That can be any of the day of a week.

So, Total possible days are:

(Sunday, Monday)

(Monday, Tuesday)

(Tuesday, Wednesday)

(Wednesday, Thursday)

(Thursday, Friday)

(Friday, Saturday)

(Saturday, Sunday)

So, here favorable outcome = 2 {i.e.(Sunday, Monday), (Saturday, Sunday)}

Total possible outcomes = 7 (written above)

$\bf \huge \fbox \blue{ \: \: required \: probability \: is \: \frac{7}{2} \: \: }$