# Statistics Question?

For Events J and K P(J)=0.4 P(J|K)=0.64

P(K|J)=0.2

a) Find P(J U K)

b) Find P(J'nK')

Relevance
• PART A:

Conditional probability says:

P(A | B) = P(A ∩ B) / P(B)

P(K | J) = P(K ∩ J) / P(J)

Plug in your known values and we have:

0.2 = P(K ∩ J) / 0.4

0.08 = P(K ∩ J)

We can turn that around and we have:

P(J ∩ K) = 0.08

But that's not what we wanted. Just use another rule of probability:

P(A ⋃ B) = P(A) + P(B) - P(A ∩ B)

In your case:P(J ⋃ K) = P(J) + P(K) - P(J ∩ K)

= 0.4 + P(K) - 0.08

= 0.32 + P(K)

Bayes' Law says:

P(A | B) = P(B | A) * P(A) / P(B)

So now we should have what we need:

P(J | K) = P(K | J) * P(J) / P(K)

0.64 = 0.2 * 0.4 / P(K)

0.64 = 0.08 / P(K)

P(K) = 0.08 / 0.64

P(K) = 8/64 = 1/8 = 0.125

Finally put that together from your result before:

P(J ⋃ K) = 0.32 + P(K)

P(J ⋃ K) = 0.32 + 0.125

P(J ⋃ K) = 0.445

PART B:

De Morgan's Laws say:

J ⋃ K = (J' ∩ K')'

So J' ∩ K' = (J ⋃ K)'

So just take the complement of your given probability:

P(J' ∩ K') = 1 - P(J ⋃ K)

P(J' ∩ K') = 1 - 0.445

P(J' ∩ K') = 0.555

• Login to reply the answers