The probability that a computer company will get a computer hardware contract is 3/4 and the probability that it will not get a software contract is 5/9. If the probability of getting at least one contract is 4/5, what is the probability that it will get both the contracts?
Choose the most appropriate option:
A) 11/30  B) 31/60  C) 41/80  D) 51/120  E) 71/180
Let, A ≡ event of getting hardware contract   B ≡ event of getting software contract   AB ≡ event of getting both hardware and software contract.   P(A) = 3/4, P(~B) =5/9   =&gt; P(B) = 1- (5/9) = 4/9.   A and B are not mutually exclusive events but independent events. So,   P(at least one of A and B ) = P(A) + P(B) - P(AB).   =&gt; 4/5 = (3/4) + (4/9) - P(AB).   =&gt; P(AB) = 71/180.   Hence, the required probability is 71/180.   The answer is 71/180.
<result>E</result>
The probability that a computer company will get a computer hardware contract is 3/4 and the probability that it will not get a software contract is 5/9. If the probability of getting at least one contract is 4/5, what is the probability that it will get both the contracts?