This discussion is now closed.

Check out other Related discussions

Probability Question: What is the value of P(A u B')?

I am stuck on the following question and would be really grateful if someone could help me out:

1) Events A and B are such that: P(A) = 0.4; P(B) = 0.5; P(A n B) = 0.3; and P(A u B) = 0.6.

i) What is the value of P(A u B')? (Where B' is the compliment of B.)

Reply 1

Swayum

Sorry but I'm not sure what you mean by "compliment". Presuming you mean P(B') = 1 - P(B):

P(AuB') = P(AnB')/P(B')

P(B') = 1 - P(B) = 0.5

P(AnB') = P(A) - P(AnB) (this is visible from a Venn diagram) = 0.1

P(AuB') = P(AnB')/P(B')
= 0.1/0.5 = 0.2.

Could be wrong, I'm not very good at probability.

*Edit*

Ignore the nonsense here.

Reply 2

datr

Here's the venn diagram. The shaded region represents the probability P(AuB'). Can you see what the answer is now?

DeathAwaitsU you have found P(A|B').

venn.gif8.6KB

Reply 3

generalebriety

DeathAwaitsU

Sorry but I'm not sure what you mean by "compliment". Presuming you mean P(B') = 1 - P(B)

That's what complement means, yes. :wink:

Compliment means B' is saying nice things to B.

Reply 4

Swayum

Oh right yeah of course, what was I thinking?

I meant:

P(AuB') = P(A) + P(B') - P(AnB')
= 0.4 + 0.5 - 0.1 = 0.8.

Reply 5

PeterParadox

DeathAwaitsU

Oh right yeah of course, what was I thinking?

I meant:

P(AuB') = P(A) + P(B') - P(AnB')
= 0.4 + 0.5 - 0.1 = 0.8.

How did you get 0.1 for P(AnB')?

Reply 6

generalebriety

PeterParadox

How did you get 0.1 for P(AnB')?

Datr explained it for you with a diagram.

Reply 7

nota bene

P(A)-P(A n B)=P(A n B')

Reply 8

steve2005

nota bene

P(A)-P(A n B)=P(A n B')

I don't think this is correct. I might be wrong, but it looks odd?

Reply 9

Gaz031

generalebriety

That's what complement means, yes. :wink:

Compliment means B' is saying nice things to B.

That made me laugh out loud :smile:

. Presumably anyone in my house who isn't fast asleep now thinks I'm crazy.

Reply 10

onkiratu

between 0.8 and 0.2, which is correct for P(AuB')

Reply 11

crax

P(AuB)=P(A) P(B)-P(A intersection B)
P(AuB')=P(A) P(B')-P(A intersection B')
P(B')=1-P(B) = 1-0.5= 0.5
P(A intersection B') = P(A).P(B') = 0.4*0.5 = 0.2
NOW:
P(AuB') = 0.4 0.5 - 0.2 = 0.7

(edited 4 years ago)