Question

It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?

Answer 1

SOLUTION

Let E be the event of having the same birthday.P(E) = 0.992⇒ P(E) + P(not E) = 1⇒ P(not E) = 1 – P(E)⇒ 1 - 0.992 = 0.008The probability that the 2 students have the same birthday is 0.008

Answer 2

Answer:

The probability that the 2 students have the same birthday is 0.008.

Step-by-step explanation:

Given : It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992.

To find : What is the probability that the 2 students have the same birthday?

Solution :

We know that, The sum of the probabilities is 1.

Let the probability of students not having the same birthday be P(E),

Then the other be P(E')

We have given, P(E)=0.992

$P(E)+P(E')=1$

$0.992+P(E')=1$

$P(E')=1-0.992$

$P(E')=0.008$

Therefore, The probability that the 2 students have the same birthday is 0.008.