Question

Find the probability that a randomly chosen group of 9 people have no birthdays in common, but randomly chosen person #10 has a birthday in common with one of the first 9.

Answer 1

Let us find the probability that no ones birthday matches among 9 people.

Probability is given by

Proabability = Favorable number of ways / total number of ways

Number of ways one person have birthday is 365

Total number of ways 9 people have birthdays is 3659

Now we calculate favorable number of ways in which no one shares their birthday

Now the number of ways first person out of 9 have birthday is 365 , second person will have 364 choices , third person will have 363 choises and .... 9 th person will have (653-8) ways

Thus, Favorable numer of ways = 365 ! / (365-9)!

(Note : This can be explained with a small example

Let us consider 3 people in place of 9 people

Now the number of ways first person have birthday is 365 , second person will have 364 choices , third person will have 363 choises

combining we have number of ways 3 person have different birthdays = 365*364*363

this can be written as

365! / (365-3)! = 365! / 362! = 365*364*363 *362! / 362! = 365*364*363 )

Thus , Probability that randomly chosen 9 people have no birthdays in common

= 0.9054

Now, we have to find that randomly chosen 10th person have common birthday with one of the first 9

9 person have 9 different birthday dates out of 365 days

Number of ways one person have birthday is 365 , this is total number of ways

Favorable number of ways = 9 (If the person has any one the 9 dates as his birthday , then his birthday matches with one of the 9 persons)

Thus probability that randomly chosen 10th person have common birthday with one of the first 9

= 9/365

= 0.0247