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Mathematics 15 Online
OpenStudy (anonymous):

Given the following: p(a)= .72 p(b)=.28 p(a n B)=.14 Find each of the following. a) the probability of A or BThis is a notation for the probability of A or B. b) the probability of the complement of AThis is a notation for the probability of the complement of A. c) the probability of the complement of B

OpenStudy (anonymous):

@Loser66 please help

OpenStudy (anonymous):

similar to the last one

OpenStudy (anonymous):

@satellite73 you must be tired of helping me. thank you

OpenStudy (anonymous):

\[P(A\cup B)=P(A)+P(B)-P(A\cap B\]

OpenStudy (anonymous):

nah not at all

OpenStudy (anonymous):

let me know what you get

OpenStudy (anonymous):

okay one sec, i'm a slow processor

OpenStudy (anonymous):

is it clear what you have to do? add \(P(A)\) to \(P(B)\) then subtract \(P(A\cap B)\) from the result

OpenStudy (anonymous):

yes I understand that

OpenStudy (anonymous):

.86?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

yay!

OpenStudy (anonymous):

second one is easier

OpenStudy (anonymous):

for that one, \(P(A^c)=1-P(A)\)

OpenStudy (anonymous):

.28

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

and the last one?

OpenStudy (anonymous):

hehe there are 3 more after the last one, would you mind helping me with them as well?

OpenStudy (anonymous):

no i don't mind at all, but first tell me what you got for the probability of B compliment

OpenStudy (anonymous):

.72 for number 3

OpenStudy (anonymous):

ok good

OpenStudy (anonymous):

the probability of the complement of A and the complement of BThis is a notation for the probability of the complement of A and the complement of B (or neither A nor B).

OpenStudy (anonymous):

wait that would be .14 right?

OpenStudy (anonymous):

hold on you last me for a second

OpenStudy (anonymous):

the probability of the complement of A and the complement of B (or neither A nor B).

OpenStudy (anonymous):

neither A nor B? ok

OpenStudy (anonymous):

is that right? .14

OpenStudy (anonymous):

that is the compliment of \(A\cup B\)

OpenStudy (anonymous):

so first we compute \(P(A\cup B)\) which you already did and got \(.86\) then you subtract it from 1 and get \(.14\)

OpenStudy (anonymous):

yeah looks like you got it !

OpenStudy (anonymous):

so do I subtract .14 from one and get .86

OpenStudy (anonymous):

wait no?

OpenStudy (anonymous):

no you did it right you subtract \(1-.86\) and get \(.14\) that is the correct answer

OpenStudy (anonymous):

oh okay. sorry I was confused for a sec.

OpenStudy (anonymous):

the next one is p(a n b^c)

OpenStudy (anonymous):

does that make sense?

OpenStudy (anonymous):

ok lets go to the picture

OpenStudy (anonymous):

okay!

OpenStudy (anonymous):

|dw:1373855781152:dw|

OpenStudy (anonymous):

would you tpe in words what the question is asking and I can figure it out

OpenStudy (anonymous):

now \(A\cap B^c\) is everything that is in A and that is NOT in B

OpenStudy (anonymous):

okay would it be .58

OpenStudy (anonymous):

do you understand what i wrote there?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

yes I understand it

OpenStudy (anonymous):

looks like you got this

OpenStudy (anonymous):

any more?

OpenStudy (anonymous):

one more! thank you so much you are a lifesaver

OpenStudy (anonymous):

no problem

OpenStudy (anonymous):

it's p(a^c n b) so I think its everything that is B and NOT a?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

.14! thank you so much! I understand. Pictures help a lot. And if you keep seeing me on here don't be alarmed, I have like 10 assignments due by midnight. thanks for all your help

OpenStudy (anonymous):

better get busy is it all probability?

OpenStudy (anonymous):

the next two assignments are, then it becomes functions

OpenStudy (anonymous):

nothing like waiting until the last minute...

OpenStudy (anonymous):

yay?(: haha

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