GATE IT 2008 | Question: 23

Question

What is the probability that in a randomly chosen group of $r$ people at least three people have the same birthday?

• A. $1-\dfrac{365-364 \dots (365-r+1)}{365^{r}}$
• B. $\dfrac{365 \cdot 364 \dots (365-r+1)}{365^r}+ ^{r}C_{1}\cdot 365 \cdot \dfrac{364.363 \dots (364-(r-2)+1)}{364^{r+2} }$
• C. $1- \dfrac{365 \cdot 364 \dots (365-r+1)}{365^{r}} - ^{r}C_{2} \cdot 365 \cdot \dfrac{ 364 \cdot 363 \dots (364-(r-2)+1)}{364^{r-2}}$
• D. $\dfrac{365 \cdot 364 \dots (365-r+1)}{365^{r}}$

Comments

self note-

suppose there are only 3 days in an year – 1,2,3

group has 4 people – r=4

then a situation with bdays like (1,1,2,2) is valid bcz person1 has same bday with p2 , p2 has same bday with p1 , p3-p4 and p4-p3.

so total 4 person have same birthdays which is >= at least 3 person having same birthdays

Answer 1

Probability of 3 or more people having birthday on same day = 1 – (probability of 2 people having birthday on same day + probability of all having distinct birthdays)

All possible ways to choose a birthday for r people = 365*365*… (r times) = $365^{r}$

Case 1: All have distinct birthdays.

No of ways = $^{365}P_{r}$.

Probability = $\frac{^{365}P_{r}}{365^{r}}$
 

Case 2: Among 365 people If exactly 2 persons have birthdays on same day.

Then we can consider these 2 people as a unit.

Ways to choose 2 people out of r people = $^{r}C_{2}$

Then the birthday for the unit of these two people can be chosen in 365  ways.

For the remaining r-2 people, we can choose distinct birthdays out of the remaining 364 days in $^{364}P_{r-2}$ ways.

Probability = $\frac{^{r}C_{2} * 365 * ^{364}P_{r-2}}{365^{r}}$
 

Hence, Probability of 3 or more people having birthday on same day

= 1 – ( $\frac{^{365}P_{r}}{365^{r}}$ + $\frac{^{r}C_{2} * 365 * ^{364}P_{r-2}}{365^{r}}$ )

You can try out the above formula with r=3. The expected answer is $\frac{1}{365^{2}}$

No option matches this. Option C may be a typo.