Question

You have a large bag in front of you and randomly pick out one candy. Find P(orange), the probability that you select a orange candy. Orange: ?, Yellow: 0.30, Brown: 0.40

Answer 1

turns out the answer was 0.30

Answer 2

i have another one to

Answer 3

I can try to answer it

Answer 4

Find P(X>2). 1: 0.30, 2: 0.40. 3: 0.15, 4: ?

Answer 5

Ohhh...um.....yeah I don't know how to do that sorry...

Answer 6

Which events are greater than two? Then, sum them up. P[X=3 or X=4].

Answer 7

would it be 0.30?

Answer 8

The reason the previous answer was 0.3 is this: if the events are mutually exclusive and exhaustive, then realize that the probability of *something* happening (one of the events) is necessarily 1. Then, P[orange] + P[yellow] + P[brown] = 1.

Answer 9

Yes, P[X=4] must be 0.15. Thus, since these events are mutually exclusive, P[X=3 or X=4] = P[X=3] + P[X=4].

Answer 10

Which is, as you indicated, 0.15+0.15 = 0.3.

Answer 11

your a life saver. thanks so much

Answer 12

Another way to answer that question: P[X>2] = 1-P[X<=2]

Answer 13

P[X<=2] -> P[X=1 or X=2] = P[X=1] + P[X=2] (since they are mutually exclusive events)

Answer 14

what if 1: 0.30, 2: 0.40, 3: 0.15, 4: 0.15 and i have to find P( not a 3)

Answer 15

P[X is not 3] = 1 - P[X=3]

Or, alternatively: P[X is not 3] = P[X=1] + P[X=2] + P[X=4]

Answer 16

so... the answer is?