Events $E_{1}, E_{2}, E_{3}$ are possible events of an experiment and their probabilities are recorded.Mark the possible correct answers.

P (E_{1}) = 0.3, P (E_{2}) = 0.4, P (E_{3}) = 0.3

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P (E_{1}) = 0.1, P (E_{2}) = 0.4, P (E_{3}) = 0.5

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P (E_{1}) = 0.6, P (E_{2}) = - 0.3, P (E_{3}) = 0.7

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P (E_{1}) = 0.4, P (E_{2}) = 0.2, P (E_{3}) = 0.3

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Solution

The correct options are
A $P (E_{1}) = 0.3, P (E_{2}) = 0.4, P (E_{3}) = 0.3$
C $P (E_{1}) = 0.1, P (E_{2}) = 0.4, P (E_{3}) = 0.5$
Since $E_{1}$ , $E_{2}$ and $E_{3}$ are mutually exclusive and exhaustive events of the experiment, the sum of the probabilities of the three events must be equal to $1$ . Also, the probability of an event cannot be negative.

In option $C$ , probability of event $2$ is negative, i.e $P (E_{2}) < 0$ . Hence option $C$ cannot be correct.
In option $D$ , sum of probabilities $P (E_{1}) + P (E_{2}) + P (E_{3}) = 0.9 \neq 1$ . Hence option $D$ cannot be correct.

Since options, $A$ and $B$ satisfy both criteria.
therefore the correct answer is $(A), (B) .$

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