Question Find $P(E_3)$ given $P(E_1) = P(E_2) = P(E_4) = P(E_5) = 0.1$ and $E_1, E_2, E_3, E_4, E_5$ are mutually exclusive events.

Studdy Solution
Calculate the probability of $E_3$ .
$P(E_3) = 0.6$
The probability of event $E_3$ is $0.6$ .

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