Question

what is the probability that two friends have different birthdays *​

Answer 1

Answer:

The probability that two different friend have different birthdays (ignoring a leap year) is : a) 364/365.

Step-by-step explanation:

Here's how you can find the solution:

Let the probability of two friends having their birthdays on any one day of the year be P.

The probability of the 1st friend having the birthday on any day of the year would be a sure event, P1= 365/365

The probability of the 2nd friend having the birthday on the same day as 1st friend’s, P2= 1/365

The probability of both the events happening simultaneously would be, P= P1 * P2 = 1/365

Hence, the probability of them having different birth dates (P’) = 1–(1/365) = 364/365

But, consider the case where the problem is to find the probability of two friends having their birthday on a given day of the year (say 21st January)

Here. the probability of the 1st friend having the birthday on 21st Jan (P1)= 1/365

The probability of the 2nd friend having the birthday on the same day as 1st friend’s birthday (P2)= 1/365

So, the probability of them having the birthday on the same day (P) = P1 * P2 = (1/365)*(1/365) = (1/365)^2

The Probability of them having different birth dates (P’) = 1–(1/365)^2

The 1st friend may be born on any 365 days of the year and similarly the 2nd friend.

Hence, total number of possible ways in which the 2 Friends have birthdays is( 365×365).

As regards the number of favourable cases out of these, we note that the 1st friend may have any of the 365 days of the year as its birthday. Similarly, the 2nd friend should be born on any of the remaining 364 days.

So, the number of cases favourable to the event “different birthdays” is( 365×364).

Now,. P = (365×364)/(365×365) = 0.997 [Ans]