each of the integers from 0 to 9 , inclusive , is written on a separate slip of blank paper and the ten slips are dropped into a hat . if 5 of the slips are the drawn , without replacement , what is the probability that all 5 have a odd number written on it ?
"key is that there is no replacement , so each successive choice will become more skewed towards picking a neg ( i . e . the pool of positives decreases , while the pool of negatives stay the same ) p ( + on 1 st pick ) = 5 / 10 p ( + on 2 nd pick ) = 4 / 9 p ( + on 3 rd pick ) = 3 / 8 p ( + on 4 rd pick ) = 2 / 7 p ( + on 5 rd pick ) = 1 / 6 5 / 10 * 4 / 9 * 3 / 8 * 2 / 7 * 1 / 6 = 1 / 252 a"
multiply(multiply(multiply(divide(add(5, const_1), add(9, const_1)), divide(subtract(add(5, const_1), const_1), subtract(add(9, const_1), const_1))), divide(subtract(subtract(add(5, const_1), const_1), const_1), subtract(subtract(add(9, const_1), const_1), const_1))), divide(subtract(subtract(subtract(add(5, const_1), const_1), const_1), const_1), subtract(subtract(subtract(add(9, const_1), const_1), const_1), const_1)))