P(3 out of 5 dice show a 3)

Question

Recently, I was running through some complementary counting and casework problems when I found this one.

"Let's say you have 5 fair six sided dice. You roll all of them. What's the probability that at least three of the dice show a 3?"

Here's my thinking:

Ways to get 3 (3s): This is $6 \cdot 6 = 36$. Ways to get 4 (3s): This is just $6$. Ways to get 5 (3s): This is $1$.

Adding this together and dividing by the denominator ($6^6$). This gives us $\frac {43}{7776}$. However, I feel like this is WAAAAAAAAAAAAAAAY to small to be the answer.

Can someone very my thinking (if I'm even right), or can anyone correct me. I'd really love some help!!

Answer 1

Note that the complement of this is the probability at most two dice out of the five show a three.

Case 1: Two of the dice show a three

There are $\dbinom{5}{2} = 10$ ways of choosing two of the dice, and a probability of $\left ( \dfrac{1}{6} \right )^2 \left ( \dfrac{5}{6} \right )^3 = \dfrac{125}{7776}$ of those two dice being 3s only. Therefore, the probability of this case is $10 \times \dfrac{125}{7776} = \dfrac{1250}{7776}$.

Case 2: One of the dice shows a three

There are $\dbinom{5}{1} = 5$ ways of choosing one of the dice, and a probability of $\left ( \dfrac{1}{6} \right ) \left ( \dfrac{5}{6} \right )^4 = \dfrac{625}{7776}$ of the one dice being a 3. Therefore, the probability of this case is $5 \times \dfrac{625}{7776} = \dfrac{3125}{7776}$.

Case 3: None of the dice show a three

There are $\dbinom{5}{0} = 1$ ways of choosing none of the dice, and a probability of $\left ( \dfrac{1}{6} \right )^0 \left ( \dfrac{5}{6} \right )^5 = \dfrac{3125}{7776}$ of none of the dice being 3s. Therefore, the probability of this case is $1 \times \dfrac{3125}{7776} = \dfrac{3125}{7776}$.

Thus, the total probability of at most two 3s is $\dfrac{1250}{7776} + \dfrac{3125}{7776} + \dfrac{3125}{7776} = \dfrac{7500}{7776} = \dfrac{625}{648}$.

Therefore, the probability there are at least three 3s is $1 - \dfrac{625}{648} = \boxed{\dfrac{23}{648}}$.

Answer 2

5C3(1/6)^3(5/6)^2= probability of getting three 3s out of five rolls.

5C4(1/6)^4(5/6)^1= probability of getting four 4s out of five rolls.

5C5(1/6)^5(5/6)^0= probability of getting all 3s.

According to the addition rule for probability, you can just add these and get the answer you want.

The way I solved is by using a formula called the binomial distribution.

nCr*(p^r)*(1-p)^(n-r)

Where r= total number of successes P= probability of a success on an individual trial n= number of trial

I thought this way is easier than counting all the probabilities.

Hope this helps! ^^