Solution:
Let E be the event wherein 2 students having the same birth date.
A/Q
Given, P(E) = 0.992
As we know,
P(E)+P(not E) = 1
So, P(not E) = 1–0.992 = 0.008
∴ The probability that the 2 students have the same birth date is 0.008
In a group of 3 students, it is given that the probability of 2 students not having the same birth date is 0.992. What is the probability that the 2 students have the same birth date?
In a group of 3 students, it is given that the probability of 2 students not having the same birth date is 0.992. What is the probability that the 2 students have the same birth date?