SOLUTION: If P(A) = 1/2,P(B) = 7/10,P(AUB) = 1,find (i) P (A ∩ B) (ii) P(A' U B')

Algebra ->  Probability-and-statistics -> SOLUTION: If P(A) = 1/2,P(B) = 7/10,P(AUB) = 1,find (i) P (A ∩ B) (ii) P(A' U B')       Log On


   

Question 1066489: If P(A) = 1/2,P(B) = 7/10,P(AUB) = 1,find
(i) P (A ∩ B) (ii) P(A' U B')


Found 2 solutions by Bay0913, 0123467:
Answer by Bay0913(3) About Me  (Show Source):
You can put this solution on YOUR website!

the first one it is asking the probability that both A&B occur. Think if that symbol as an "A" which means and. So to find the probability of both occurring at the same time, just multiply them by each other. so (1/2)*(7/10)=(7/20) That is your answer for the first one.
The second is a bit trickier and not as straight forward.
The formula for P(A' U B') is P(A)+P(B)-P(A∩B), and since the ' symbol is there, it means 1-P(A) & 1- P(B)
so you're going to do (1-1/2)+(1-7/10)-(7/20) which is
.5+.3-.35=.5
That's your answer, hope it helps.

Answer by 0123467(1) About Me  (Show Source):
You can put this solution on YOUR website!
i) P(A ∩ B)
1/2 + 7/10 - 10/10
= 5/10 + 7/10 - 10/10
=2/10
=1/5
ii) P(A' U B')
P(B')= 1-P(B)
= 1-7/10 = 3/10
P(A' ∩ B')= 0
P(A' U B') = P(A') + P(B') = P(A' ∩ B')
= 5/10 + 3/10 - 0/10
=8/10
= 4/5