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# If $e_1,e_2,e_3,e_4$ are the four elementary outcomes in a sample space and $P(e_1) = .1, P(e_2) = .5, P (e_3) = .1$, then the probability of $e_4$ is ______.

$\begin{array}{1 1}(A)\;0.3\\(B)\;0.4\\(C)\;0.5\\(D)\;0.13\end{array}$

Sum of probabilities =1
$P(e_1)+P(e_2)+P(e_3)+P(e_4)=1$
$0.1+0.5+0.1+P(e_4)=1$
$P(e_4)=1-0.7$
$P(e_4)=0.3$
Hence (A) is the correct answer.