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Answer 1

▶ Question :-

→ Find the probability that a leap year selected at random, will contain 53 Mondays .


▶ Answer :-

→ The probability of getting 53 Mondays in a leap year = 2/7 .



▶ Step-by-step explanation :-


As we know , a normal year contains 365 days and a leap year contains 366 days .


Therefore, 365 days = 52 weeks and 1 day .

And, 366 days = 52 weeks and 2 days .

So, we have to find the probability of 53 Mondays in leap years.


▶ Now,


→ 52 Mondays are compulsary as leap year contains 52 weeks .

And, the probability of remaining 2 day may be :

→ ( Sunday , Monday ) .

→ ( Monday , Tuesday ) .

→ ( Tuesday , Wednesday ) .

→ ( Wednesday , Thursday ) .

→ ( Thursday , Friday ) .

→ ( Friday , Saturday ) .

→ ( Saturday , Sunday ) .


Therefore , total number of probability of Mondays is 2 in 7 probabilities .



°•° P(E) of getting 53 Mondays in a leap year

$\huge \orange{ \boxed{ \boxed{ \sf = \frac{2}{7} .}}}$


Hence, it's solved .

Answer 2

Question:- Find the probability that a leap year selected at random, will contain 53 Mondays.

Answer:  2/7

Step-by-step explanation:

We know that  there  are  365 days  in a  year and  a  leap years contains 366 days.

There  are  7 days in  Week

366 Days = 52 weeks + 2 more day.

Now, It is  certain that  52 weeks  contains 52 Mondays surely.

But it  is  not certain what will be  remaining  2  days. It  may be as follows:-

1. Monday  and Tuesday

2. Tuesday and  Wednesday

3. Wednesday and  Thursday

4. Thursday and  Friday

5. Friday and Saturday

6. Saturday and  Sunday

7. Sunday and Monday

We can see  that  there total 7 possibilities and  in this, there are   chances  of  having  day  Monday is  2 ( in case 1 and 7)

We have, Total Possible Outcomes = 7

Total Favourable Outcome = 2

Probability = Total Favourable Outcome/Total Possible Outcomes

                  = 2/7