Question

Two friends were born in the year $2000$. The probability that they have the same birth date is
A) $\dfrac{1}{{2000}}$
B) $\dfrac{2}{{365}}$
C) $\dfrac{1}{{365}}$
D) $\dfrac{1}{{366}}$

Answer 1

Hint:Total number of days in the year $2000 = 366$ days as because $2000$ was the leap year. And we know that the probability is equal to $\dfrac{{{\text{number of favourable outcome}}}}{{{\text{total number of outcome}}}}$.So the number of favorable outcomes here is $1$ as both can have only one day as both have birthdays on the same day.Using these definitions and concepts we try to get the answer.

Complete step-by-step answer:
As we are needed to find the probability that two friends which were born in the year $2000$ have the same birth date. So for this we need to know that in the year $2000$ how many days were there as we know that the year which is an integral multiple of $4$ is termed as leap year and normal year contains $365$ but leap year contains $366$ days. Here it is given in the year $2000$. So we need to find whether it is leap year or not.
When we divide $2000$ by $4$ i.e. $\dfrac{{2000}}{4} = 500$
We get an integer. Therefore $2000$ is the integral multiple of $4$ therefore It is the leap year and as we know leap year has $366$ days and we are given that two friends born in year $2000$ have the same birth date.
So number of favourable outcomes would be that there birthday can be any day of $366$ days that means favorable outcomes for their birthday will be $1$ and total number of outcomes in the year $2000$ will be $366$
So, Probability that both have same birthday $ = \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}} = \dfrac{1}{{366}}$

So, the correct answer is “Option D”.

Note:We know that the probability of something which is impossible to happen will be $0$ & the probability of any event that has a chance to happen lies in the range $[0,1]$ .For any event P(Event Happening)=1-P(Event Not Happening).