Find the missing probability in the table below.
X 1 2 3 4
P(X) 0.40 0.20 0.20 ?
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jimthompson5910 (jim_thompson5910):
All probabilities must add to 1 in a probability distribution
jimthompson5910 (jim_thompson5910):
So if y is the missing probability, then
0.4+0.2+0.2+y = 1
OpenStudy (anonymous):
I don't understand..
jimthompson5910 (jim_thompson5910):
add up 0.4, 0.2, and 0.2
OpenStudy (anonymous):
0.8?
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jimthompson5910 (jim_thompson5910):
so there's an 80% chance that X is either 1, 2, or 3
jimthompson5910 (jim_thompson5910):
what's left over?
OpenStudy (anonymous):
20%? I think
jimthompson5910 (jim_thompson5910):
yes, there's a 20% chance of X = 4 since that's the only thing left
jimthompson5910 (jim_thompson5910):
so the missing probability is 0.2
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OpenStudy (anonymous):
0.2! Got it thanks ! (:
OpenStudy (anonymous):
Can i bother you with another question?
jimthompson5910 (jim_thompson5910):
You can't have anything greater than 20% since you can't go over 100%
and you can't have anything under 20% because if you fall short of 100%, then that suggests there is another possibility....when there clearly isn't
so it has to be 20%
jimthompson5910 (jim_thompson5910):
sure go for it
OpenStudy (anonymous):
Given the table of M&M® distributions below, find P(not orange), the probability of selecting anything but a orange M&M at random from a large bag.
Red Orange Yellow Green Blue Brown
0.13 0.20 ? 0.16 0.24 0.13
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jimthompson5910 (jim_thompson5910):
those are all the possibilities and you can't choose anything else
eg: you have a 13% chance of choosing Red. You have a 24% chance of choosing blue
jimthompson5910 (jim_thompson5910):
so add up all the given probabilities
OpenStudy (anonymous):
OKay hold onn
OpenStudy (anonymous):
0.86
OpenStudy (anonymous):
So i would subtract that from .1?
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jimthompson5910 (jim_thompson5910):
so the missing probability must be what
jimthompson5910 (jim_thompson5910):
from 1 not 0.1
OpenStudy (anonymous):
Is .14 the answer?
jimthompson5910 (jim_thompson5910):
no, but it helps get you there
OpenStudy (anonymous):
Shoot.
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jimthompson5910 (jim_thompson5910):
they want to know the probability of choosing anything but orange
jimthompson5910 (jim_thompson5910):
so add up all the probabilities that correspond to anything but orange
jimthompson5910 (jim_thompson5910):
so add the probabilities for red, yellow, green, blue, brown
OpenStudy (anonymous):
Okay i got 0.8
OpenStudy (anonymous):
Which is 80%
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jimthompson5910 (jim_thompson5910):
good, so the answer is 0.8
OpenStudy (anonymous):
A thousand times thank you!
jimthompson5910 (jim_thompson5910):
which corresponds to an 80% chance of selecting something that isn't orange