Question

7. 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 sam
birthday?
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Answer 1

Step-by-step explanation:

answer is 0.008

because p(E)+p(Ebar)=1

so,p(Ebar)+0.992=1

p(Ebar)=0.008

Answer 2

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

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