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 3/5. If the probability of getting at least one contract is 4/5, what is the probability that it will get both the contracts? Pick one:
A) 4/11
B) 5/12
C) 6/17
D) 7/20
E) 11/30
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- (3/5) = 2/5.   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) + (2/5) - P(AB).   =&gt; P(AB) = 7/20.   Hence, the required probability is 7/20.   The answer is 7/20.
<result>D</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 3/5. If the probability of getting at least one contract is 4/5, what is the probability that it will get both the contracts?