Probability of all elements question

[mutzy188]

Homework Statement

If P(A) =0.4, P(B)=0.5, and P(A∪B)=0.7, find P(A’∪B’)

The Attempt at a Solution

P(A’) = (Probability of all elements in S that are not in A) = 1 - P(A) = 0.6

P(B’) = (Probability of all elements in S that are not in B) = 1 – P(B) = 0.5

P(A’∪B’)= The union of A’ and B’ = 1 - 0.7 = 0.3

P(A’∪B’) = P(A’) + P(B’) – P(A’∩ B’) = 0.6 + 0.5 – 0.3 = 0.8

So, P(A’∪B’) = 0.8

I'm not sure if i did this correctly.

Thanks

 

[EnumaElish]

How do you know P(A’∩ B’) = 0.3?

The easiest method is to figure P(A ∩ B). (Can you tell me why?)

 

[mutzy188]

I just assumed that you could do that.

P(A ∩ B) = .4 + .5 - .7 = .2

So would P(A' ∩ B') = 1 - .2 = .8 ?

 

[wbclark]

You're correct but you seem to be bent on making this too difficult =P. P(A intersect B) = P(A) * P(B) = 0.4 * 0.5 = 0.2

I know you probably get this all the time, but drawing pictures for these things really does help. Or even just visualizing a Venn Diagram for these events in your head. You can see that the complement of (A' union B') is (A intersect B) and P(A intersect B) is easy to find.

 

[EnumaElish]

P(A ∩ B) = .4 + .5 - .7 = .2

Correct; and that's exactly how I would have calculated this.

So would P(A' ∩ B') = 1 - .2 = .8 ?

No. P(A ∩ B)' = 1 - .2 = .8 is correct. What is the relationship between (A ∩ B)' and A' U B' ?