Let $E_{1}, E_{2}, E_{3} . . . E_{N}$ be the 'n' exhaustive events in a random experiment then. $p (E_{1}) p (, E_{2}), p (E_{3}) . . . p (E_{N})$ is

less than

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equal to

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greater than 1

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none

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Solution

The correct option is D none
We have,

A set of events is said to be exhaustive events if these are mutually exclusive as well as exhaustive events.

Events $E_{1}, E_{2}, E_{3} . . . . . . . . . . . E_{n}$ are such that, $E_{1} \cap E_{2} \cap E_{3} \cap . . . . . . . . . . \cap E_{n} = ϕ$

Then these events are known as exhaustive events.
hence, $P (E_{1}) P (E_{2}) P (E_{3}) . . . . . . . . . . P (E_{n}) = n o n e$

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Let E1,E2,E3...EN be the 'n' exhaustive events in a random experiment then. p(E1)p(,E2),p(E3)...p(EN) is

Let $E_{1}, E_{2}, E_{3} . . . E_{N}$ be the 'n' exhaustive events in a random experiment then. $p (E_{1}) p (, E_{2}), p (E_{3}) . . . p (E_{N})$ is