Question

17. The probability that a non-leap year selected at random has 53 Sundays is
1/365​

Answer 1

Answer:

The probability that a non-leap year selected at random has 53 Sundays is 1/7.

Step-by-step explanation:

We know that,

A non-leap year consists of 365 days.

Now,

How many weeks are there in 365 days?

To find that out we must divide 365 and 7, because 1 week = 7 days

365 ÷ 7 = 52 and Remainder 1

Hence,

In a non-leap year, there are 52 weeks and 1 day totally.

We know that, in all the 52 weeks there will be Sundays.

Thus, 52 Sundays are a sure.

So, we now only need to find out if the last 1 day is a Sunday or not, as 52 Sundays has already gone by.

Favorable outcomes = Sunday

Number of favorable outcomes = 1

Total outcomes = Sunday, Monday, Tuesday, Wednesday, Thursday, Friday and Saturday

Number of total outcomes = 7

We know that,

Probability = No. of favorable outcomes ÷ No. of total outcomes

Probability(of 53 Sundays) = 1/7

Hence,

The probability that a non-leap year selected at random has 53 Sundays is 1/7

This answer is not wrong, it is mathematically correct, in case you feel this is wrong, you may approach your teachers or even Google it.

You may feel this is wrong naturally, this is because 1/7 compared to 53 days of Sunday to 365 is nearly different.

Let's try to solve this, we have 52 weeks and so 52 Sundays are always sure, no matter when the year starts, Now the only Question is if the last day is a Sunday or not. Thus only that last day decides if there will 53 Sundays or 52 Sundays, that's why its 1/7. We need the last day to be Sunday, but it can be any other day of a week, because the year is chosen randomly, so we conclude that it is 1/7.

Hope it helped and believing you understood it........All the best