Given: P(e1) = 0.1, P(e2) = 0.5, P(e3) = 0.1
We know that,
Sum of all probabilities = 1
∴ P(e1) + P(e2) + P(e3) + P(e4) = 1
⇒ 0.1 + 0.5 + 0.1 + P(e4) = 1
⇒ P(e4) = 1 – 0.7
⇒ P(e4) = 0.3
Ans. If e1, e2, e3, e4 are the four elementary outcomes in a sample space and P(e1) =.1, P(e2) = .5, P (e3) = .1, then the probability of e4 is 0.3