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?  Pick
A) 1/12
B) 1/10
C) 1/8
D) 1/24
E) 5/9
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 1st pick) = 5/10   P(+ on 2nd pick) = 4/9   P(+ on 3rd pick) = 3/8   P(+ on 4rd pick) = 2/7   P(+ on 5rd pick) = 1/6   5/10*4/9*3/8*2/7*1/6=1/252   1/12
<result>A</result>
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?