Question

this exercise is on probabilities and combinations.
a if two people are selected at random, the probability that they do not have the same birthday (day and month) is $\frac{365}{365}cdot\frac{364}{365}$. explain why this is so. (ignore leap years and assume 365 days in a year)
the first person can have any birthday, so they can have a birthday on (square) of the 365 days. in order for the second person to not have the same birthday they must have one of the (square) remaining birthdays. (type whole numbers)
b if four people are selected at random, find the probability that they all have different birthdays.
the probability that they all have different birthdays is (square) (round to three decimal places as needed)
c if four people are selected at random, find the probability that at least two of them have the same birthday.
the probability that at least two of them have the same birthday is (square) (round to three decimal places as needed)
d if 14 people are selected at random, find the probability that at least 2 of them have the same birthday.
the probability that at least two of them have the same birthday is (square)

Explanation:

Step1: Analyze first - person's birthday in part a

The first person can have a birthday on 365 of the 365 days.

Step2: Analyze second - person's birthday in part a

The second person must have one of the 364 remaining birthdays.

Step3: Calculate probability for 4 people having different birthdays in part b

The first person can have any of 365 birthdays. The second person has 364 non - matching birthdays, the third person has 363 non - matching birthdays and the fourth person has 362 non - matching birthdays. The probability $P(\text{4 different})=\frac{365}{365}\times\frac{364}{365}\times\frac{363}{365}\times\frac{362}{365}=\frac{365\times364\times363\times362}{365^4}\approx0.986$.

Step4: Calculate probability of at least two having the same birthday for 4 people in part c

The probability of at least two having the same birthday is the complement of all having different birthdays. So $P(\text{at least 2 same}) = 1 - P(\text{4 different})=1 - 0.986 = 0.014$.

Step5: Calculate probability of all different birthdays for 14 people

The probability that 14 people have different birthdays is $P=\frac{365}{365}\times\frac{364}{365}\times\cdots\times\frac{365 - 13}{365}=\frac{365!}{365^{14}(365 - 14)!}$.

Step6: Calculate probability of at least two having the same birthday for 14 people

$P(\text{at least 2 same})=1 - P(\text{14 different})$.
$P(\text{14 different})=\frac{365\times364\times\cdots\times352}{365^{14}}\approx0.788$.
So $P(\text{at least 2 same})=1 - 0.788 = 0.212$.

Answer:

a. First person: 365; Second person: 364
b. 0.986
c. 0.014
d. 0.212