Ten $6$-sided dice are rolled. What is the probability that exactly three of the dice show an odd number? Express your answer as a decimal rounded to the nearest thousandth.
We have a binomial probabiity problem....3 of the dice must show odd and each one has a probability of (1/2) of that....the other 7 each have a probability of (1/2) of showing even.....we need to choose any 3 of the 10 to show odd...so we have
C(10,3) (1/2)^3 (1/2)^7 = 15 /128 ≈ .117