The probability of a leap year having 53 Mondays is:
A. $\dfrac{2}{7}$
B. $\dfrac{1}{7}$
C. $\dfrac{3}{7}$
D. $\dfrac{4}{7}$
Answer
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Hint: First of all we should know how many days a leap year have, how many weeks a year have and how many arrangements are possible of weeks in a year.
Complete step by step answer:
So, here we know about a leap year:
Total number of days = 366
Total number of full weeks = 52
And, remaining days $=366-52\times 7=2$
Now, we will think that a year can start with any day out of 7 days of a week.
So, here we make different possibilities of remaining 2 days.
If leap year starts Monday then the remaining 2 days will be Tuesday and Wednesday.
Same with Tuesday: remaining 2 days Wednesday and Thursday.
Wednesday: remaining 2 days Thursday and Friday.
Thursday: remaining 2 days Friday and Saturday.
Friday: remaining 2 days Saturday and Sunday.
Saturday: remaining 2 days Sunday and Monday.
Sunday: remaining 2 days Monday and Tuesday.
Because we know 52 Mondays counted in total for a full week in a year, so we need 1 more Monday which we have to take from the remaining 2 days.
And, we know total possible cases = 7.
2 cases which have Monday in the remaining 2 day = 2.
Using direct formula $P\left( A \right)=\dfrac{\text{Number of favourable outcomes}}{\text{Total number of favourable outcomes}}$
So, Probability of leap year having 53 Monday is $=\dfrac{2}{7}$ .
And option (A) is the correct answer.
Note: Don’t confuse yourself while solving this kind of problem by counting first to last week days. Sometimes these problems can confuse you in exams because it is easy but without concepts and tricks it takes more time.
Complete step by step answer:
So, here we know about a leap year:
Total number of days = 366
Total number of full weeks = 52
And, remaining days $=366-52\times 7=2$
Now, we will think that a year can start with any day out of 7 days of a week.
So, here we make different possibilities of remaining 2 days.
If leap year starts Monday then the remaining 2 days will be Tuesday and Wednesday.
Same with Tuesday: remaining 2 days Wednesday and Thursday.
Wednesday: remaining 2 days Thursday and Friday.
Thursday: remaining 2 days Friday and Saturday.
Friday: remaining 2 days Saturday and Sunday.
Saturday: remaining 2 days Sunday and Monday.
Sunday: remaining 2 days Monday and Tuesday.
Because we know 52 Mondays counted in total for a full week in a year, so we need 1 more Monday which we have to take from the remaining 2 days.
And, we know total possible cases = 7.
2 cases which have Monday in the remaining 2 day = 2.
Using direct formula $P\left( A \right)=\dfrac{\text{Number of favourable outcomes}}{\text{Total number of favourable outcomes}}$
So, Probability of leap year having 53 Monday is $=\dfrac{2}{7}$ .
And option (A) is the correct answer.
Note: Don’t confuse yourself while solving this kind of problem by counting first to last week days. Sometimes these problems can confuse you in exams because it is easy but without concepts and tricks it takes more time.
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