Question

Suppose we pick three people at random. For each of the 2.32 The following questions, ignore the special case where someone might be born on February 29th, and assume that births are evenly distributed throughout the year. (a) What is the probability that the first two people share a birthday? (b) What is the probability that at least two people share a birthday?

Answer 1

Answer:

Step-by-step explanation:

Given that we pick up three people at random.

We can ignore the leap year so we assume there are 365 days in a year.

Probability for a person to be born on a specific data is equally likely and hence equal to

$\frac{1}{365}$

Each person out of three persons is independent of the other.

Hence for the second person to have the same birthday would have equal probability.

X = No of persons to have common birthday is binomial with n =3 and p = $\frac{1}{365}$

a) the probability that the first two people share a birthday

= $(\frac{1}{365})^2\\= \frac{1}{133225}$

b) the probability that at least two people share a birthday

= $P(X\geq 2)\\=P(x=2)+P(x=3)\\= 3C2 (\frac{1}{365} )^2*\frac{364}{365}+(\frac{1}{365})^3$