Question

It is given that the group of three student, the probability of 2 Students not having the same birthday is 0.992. What is the probability that the 2 student have same birth

Answer 1

Step-by-step explanation:

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')

(E) = 0.992

⇒ P(E) + P(E') = 1

⇒ P(E') = 1 – P(E)

⇒ 1 - 0.992 = 0.008

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

Answer 2

$heyy \: mate \\ \\ here \: is \: your \: answer \:$

let E be the event of having the same birthday

P(E) =0.992

But P(E) +P = 1

P= 1-P(E)

= 1-P(E)

=1-0.992

=0.008

$hope \: it \: is \: helpful \: for \: you$