The probability that a computer company will get a computer hardware contract is 4/5 and the probability that it will not get a software contract is 3/5. If the probability of getting at least one contract is 9/10, what is the probability that it will get both the contracts?
Select:  A) 1/8  B) 2/9  C) 3/10  D) 4/11  E) 5/12
Let, A ≡ event of getting hardware contract   B ≡ event of getting software contract   AB ≡ event of getting both hardware and software contract.   P(A) = 4/5, 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; 9/10 = (4/5) + (2/5) - P(AB).   =&gt; P(AB) = 3/10.   Hence, the required probability is 3/10.   The answer is 3/10.
<result>C</result>
The probability that a computer company will get a computer hardware contract is 4/5 and the probability that it will not get a software contract is 3/5. If the probability of getting at least one contract is 9/10, what is the probability that it will get both the contracts?