Question

concept of probability
Let P(Z)=.40, P(Y)=.30, and P(Z∪Y)=.48 Find each of the given probabilities.

a.P(Z'∩Y')
b.P(Z'∪Y')
c.P(Z'∪Y)
d,P(Z∩Y') @terenzreignz

Answer 1

Things you need to know...for any event X
P(X') = 1 - P(X)

Answer 2

So, with that in mind, what are P(Y') and P(Z') respectively?

Answer 3

1-p(y) and 1-p(z)

Answer 4

yes... but we have numerical values for P(Y) and P(Z)

Answer 5

not to butt in, but a venn diagram would solve these in seconds

Answer 6

@satellite73
This is the second time (as far as I recall) you said "not to butt in..." and then butted in...

And most annoyingly, you always make a valid point.... fml
XD

Answer 7

Okay, @payin
Let's draw that venn diagram ^_^

Answer 8

yeah i don't really mean "not to butt in" i mean "try this it will be easy"

Answer 9

ok

Answer 10

...so...
\(P(Y \cup Z) = 0.48 \)

Answer 11

So everything inside the sets representing Y and Z must total 0.48, yes?

Answer 12

ok

Answer 13

So, what to put here?
This represents their intersection.
This should be one formula you'd know by heart...
\[\Large P(Y\cap Z) = P(Y) + P(Z) - P(Y\cup Z)\]

Answer 14

ok

Answer 15

Well, what is the probability of their intersection, then?
You already know P(Y), P(Z) and P(Y U Z)
So work out \(P(Y\cap Z)\)

Answer 16

ok

Answer 17

.60

Answer 18

i mean p(.60)

Answer 19

How did you get that?

Answer 20

pz +py-p(z-y)

Answer 21

.40+.30-(.40-.30)

Answer 22

Uhh... It's not p(z-y)
it's p(z U y)

Answer 23

oh ok

Answer 24

So, fix it...

Answer 25

0

Answer 26

Why?

Answer 27

oh sorry...70-.48=.22

Answer 28

Better. So that's what we put here...

Answer 29

So, what about the rest of Y?

Answer 30

.30-.22

Answer 31

not sure

Answer 32

It's correct.
What about the rest of Z?

Answer 33

.40-.22

Answer 34

Which is...?

Answer 35

.18

Answer 36

Now, on the outside, it has to be the rest, that is, neither Y nor Z, and and it should be