Birthday and 53 Sundays Probability | Class 10 Maths

Quick answer: For two people born in the same non-leap year, P(different birthdays) = 364/365 and P(same birthday) = 1/365. A non-leap year has 52 weeks + 1 extra day, so P(53 Sundays) = 1/7. A leap year has 52 weeks + 2 extra days, so P(53 Sundays) = 2/7.

The two rules you need

Classical probability:

$P(E) = \frac{\text{number of favourable outcomes}}{\text{total number of possible outcomes}}$

Complementary events: An event either happens or does not, so

$P(E) + P(\text{not } E) = 1$

Problem 1: Same or different birthdays (NCERT)

Two friends, Savita and Hamida, were born in the same non-leap year. What is the probability that they have (i) different birthdays, and (ii) the same birthday?

Step 1: Total outcomes.

A non-leap year has 365 days. Fix Savita's birthday on any day. Hamida's birthday can then be any of the 365 days, so there are 365 equally likely outcomes.

Step 2: Different birthdays.

Hamida's birthday must fall on any day except Savita's, so there are 365 − 1 = 364 favourable outcomes.

$P(\text{different birthdays}) = \frac{364}{365}$

This is about 99.7%.

Step 3: Same birthday.

"Same birthday" and "different birthdays" are complementary events:

$P(\text{same birthday}) = 1 - \frac{364}{365} = \frac{1}{365}$

Direct check: Only one of Hamida's 365 possible birthdays matches Savita's, so the probability is 1/365 (about 0.27%).

Answer: (i) 364/365, (ii) 1/365.

Problem 2: 53 Sundays in a non-leap year

Step 1: A non-leap year has 365 days.

$365 = 52 \times 7 + 1$

So it has 52 complete weeks and 1 extra day.

Step 2: The 52 complete weeks give exactly 52 Sundays.

Step 3: The extra day can be any of the 7 days of the week, all equally likely. The year has 53 Sundays only if that extra day is a Sunday.

$P(\text{53 Sundays in a non-leap year}) = \frac{1}{7}$

Problem 3: 53 Sundays in a leap year

Step 1: A leap year has 366 days.

$366 = 52 \times 7 + 2$

So it has 52 complete weeks and 2 extra consecutive days.

Step 2: The 2 extra days can be any of these 7 pairs:

Step 3: Two of the seven pairs contain a Sunday.

$P(\text{53 Sundays in a leap year}) = \frac{2}{7}$

Variation: 53 Sundays and 53 Mondays in a leap year

Only one pair, (Sunday, Monday), contains both days, so:

$P(\text{53 Sundays and 53 Mondays}) = \frac{1}{7}$

Variation: 53 Sundays or 53 Mondays in a leap year

The pairs containing Sunday or Monday are (Sun, Mon), (Mon, Tue) and (Sat, Sun), so the probability is 3/7.

Summary table

Exam tips

• Write the division step clearly: 365 = 52 × 7 + 1, or 366 = 52 × 7 + 2.
• For leap-year questions, list all 7 pairs of extra days. This shows the sample space and earns method marks.
• In the birthday question, say that you fix the first person's birthday. This makes the total outcomes 365, not 365 × 365.
• Use the complement rule to check your answer: the two parts must add up to 1.

Common mistakes

• Writing 2/366 or 53/366 for 53 Sundays in a leap year. The 52 weeks are fixed; only the 2 extra days matter, and they form 7 equally likely pairs.
• Counting 14 pairs or writing the extra days as two separate, unconnected days. They are consecutive.
• Giving 2/365 for the same-birthday question. Only one day matches, so it is 1/365.
• Mixing up the two parts of the birthday question: different birthdays is the likely event (364/365).

Frequently Asked Questions

What is the probability that two friends have the same birthday?

In a non-leap year it is 1/365, because only one of the 365 days matches the first friend's birthday.

What is the probability that two friends have different birthdays?

It is 364/365, since the second friend's birthday can be any of the 364 days that do not match.

What is the probability of 53 Sundays in a leap year?

It is 2/7. A leap year has 52 weeks and 2 extra consecutive days, and 2 of the 7 possible pairs contain a Sunday.

What is the probability of 53 Sundays in a non-leap year?

It is 1/7. A non-leap year has 52 weeks and 1 extra day, which is a Sunday in 1 case out of 7.

What is the probability of 53 Sundays and 53 Mondays in a leap year?

It is 1/7, because only the pair (Sunday, Monday) contains both days.