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

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