I really need help with this question...Please

Question

anonymous · Answer

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loser66 · Answer

We have formula for them, 
$$P(A \cup B) = P(A) + P (B)- P(A\cap B)$$
so just plug the numbers in, you have a)

loser66 · Answer

$P(A^c) =1- P(A)$ plug the number in, you have b)

loser66 · Answer

The same with c) but for $B^c$

loser66 · Answer

Now, you need another formula 
$(A\cup B)^c= A^c \cap B^c$
so that P (right hand side) = P (left hand side) 
but P (left hand side ) = $1- P (A\cup B)$ , then you have d)

loser66 · Answer

Actually, you can visualize the problem like this |dw:1417397730502:dw|

loser66 · Answer

Hence, probability of not A ($A^c$) is 0.28
probability of not B ($B^c$) is 0.72

loser66 · Answer

so that probability of A and not B is 0.58
it is waaaaaay easier than apply formula for e and f

anonymous · Answer

I dont get it, so what did you get for a, b, c, d, and e??

loser66 · Answer

Why don't you get? just plug the number in the formula. is it that hard? for example, for a, I give you formula |dw:1417398374885:dw|