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 ?
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 .
subtract(add(divide(3, 4), subtract(const_1, divide(3, 5))), divide(5, 6))