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 ?
"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 = > 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 ) . = > 9 / 10 = ( 4 / 5 ) + ( 2 / 5 ) - p ( ab ) . = > p ( ab ) = 3 / 10 . hence , the required probability is 3 / 10 . the answer is c ."
subtract(add(divide(const_3.0, 5), subtract(const_1, divide(4, 5))), divide(5, 10))