the probability that a computer company will get a computer hardware contract is 2 / 3 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 ?
"explanation : let , a ≡ event of getting hardware contract b ≡ event of getting software contract ab ≡ event of getting both hardware and software contract . p ( a ) = 2 / 3 , p ( ~ b ) = 5 / 9 = > 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 ) . = > 4 / 5 = ( 2 / 3 ) + ( 4 / 9 ) - p ( ab ) . = > p ( ab ) = 14 / 45 . hence , the required probability is 14 / 45 . answer : a"
subtract(add(divide(2, 3), subtract(const_1, divide(2, 9))), divide(9, 5))