If A = { $w_{1}, w_{2} . . . . . w_{n}$ }, where $w_{i}$ are the outcomes, then

P(A)= P( $w_{1}$ )+P( $w_{2}$ ) ..... P( $w_{n}$ )

True

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False

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Solution

The correct option is A
True

Probability of an event is the sum of probabilities of outcomes associated with it. If we throw a die, to find the probability of an even number turns up, we will add the probability of 2, 4 and 6 (outcomes associated with it)

$\Rightarrow$ The given statement is correct.

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Similar questions

If A = { $w_{1}, w_{2} . . . . . w_{n}$ }, where $w_{i}$ are the outcomes, then

P(A)= P( $w_{1}$ )+P( $w_{2}$ ) ..... P( $w_{n}$ )

If $A = w_{1}, w_{2} \dots . . w_{n}$ , where $w_{i}$ are the outcomes, then $P (A) = P (w_{1}) + P (w_{2}) \dots . . P (w_{n})$

If A = { $w_{1}, w_{2} . . . . . w_{n}$ }, where $w_{i}$ are the outcomes, then P(A)= P( $w_{1}$ )+P( $w_{2}$ ) ..... P( $w_{n}$ )

A = (w_{1}, w_{2} \dots . . w_{n})

w_{1}, w_{2} \dots . . w_{n}

P (A) = P (w_{1}) + P (w_{2}) \dots . . P (w_{n})

1, w, w^{2}, . . . w^{n - 1}

n^{t h}

(1 - w) (1 - w^{2}) . . . (1 - w^{n - 1}) i s

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