If A - w 1- w 2- w n - where w 1- w 2- w n are the outcomes- then P A - P w 1- P w 2 - P w n A- TrueB- False

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Random Experiment

If A = w 1, w...

Question

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

True

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

False

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

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( $w_{n}$ )

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

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

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

n^{t h}

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

Join BYJU'S Learning Program