What is the probability of getting exactly #2# three's when rolling #4# fair, #6#-sided die?
I need a solution that isn't #(4c2)*(1/6)^2*(5/6)^2#
I need a casework type answer.
I need a solution that isn't
I need a casework type answer.
1 Answer
The solution is indeed
Explanation:
This question uses a binomial distribution—repeated independent trials, each with only two options (success or failure).
For each die, the probability of getting a 3 is
We are interested in the probability of getting exactly 2 threes. This means the other two dice show anything that's NOT 3.
Let's imagine the dice are rolled one at a time. What's the probability of rolling the two 3's on the first two rolls? Since all 4 rolls are independent, this would be
#"P"(3) * "P"(3) * "P"("not 3") * "P"("not 3")#
#=1/6 * 1/6 * 5/6 * 5/6#
#=(1/6)^2(5/6)^2#
But wait—this can't be our final answer, because there are other orders in which we could roll two 3's. (i.e. 3x3x, 3xx3, etc.)
To make this right, we need to account for all the orders in which we could get two 3's out of four rolls. This is where the
33xx, 3x3x, 3xx3, x33x, x3x3, xx33
To get the total probability for two 3's in any order, we add the probabilities of each of these orders together. Since they all involve independently rolling two 3's and two NOT 3's, each order has the same probability of
This is where we get
While many probability tests will allow you to stop here (because it shows you know how to arrive at the answer), you can easily multiply this out to get a proper fraction, and then an approximate decimal:
#""_4C_2 (1/6)^2 (5/6)^2 = 6(1/36)(25/36)#
#color(white)(""_4C_2 (1/6)^2 (5/6)^2) = 25/216#
#color(white)(""_4C_2 (1/6)^2 (5/6)^2) ~~ 0.1157" "= 11.57%#