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 5/6, what is the probability that it will get both the contracts? Choose the correct choice from the following options
A) 7 / 30
B) 11 / 40
C) 17 / 50
D) 19 / 60
E) 33 / 80

Rationale: 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 = > 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 ) . = > 5 / 6 = ( 3 / 4 ) + ( 2 / 5 ) - p ( ab ) . = > p ( ab ) = 19 / 60 . hence , the required probability is 19 / 60 . the answer is d .
Answer: D