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 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 ?
"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 - ( 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 = ( 3 / 4 ) + ( 4 / 9 ) - p ( ab ) . = > p ( ab ) = 71 / 180 . hence , the required probability is 71 / 180 . the answer is e ."
subtract(add(divide(5, 4), subtract(const_1, divide(const_3.0, 4))), divide(9, 5))