[Math] Calculation probabilities for rolling 2 dice 5 times.

Question

Two fair dice are rolled 5 times. Let the random variable x represent the number of times that the sum 6 occurs. The table below describes the probability distribution. Find the value of the missing probability.

Probability table

The value I entered is correct because I know they must sum to one. I was wondering how I would calculate this without the rest of the table though. Here was my thought process which does not seem to be correct.

Possible ways to sum to 6 would be $\frac{5}{36}$. So I assumed this could then be used from a binomial probability distribution standpoint. So assuming I am looking for the sum of 6 occurring 4 times out of 5 tosses:

$ 5 \choose 4$ $*(\frac{5}{36})^{4}(1-\frac{5}{36})^{1} $

I thought the above should give me the the probability for having 4 out of my 5 tosses be successful in having the sum 6 occur.

Answer 1

You seem to be on the right track with the random variable $X \sim \mathsf{Binom}(5, 5/36).$ Maybe you are making a computational error:

From R statistical software:

x=0:5; pdf=dbinom(x,5,5/36)
cbind(x, pdf)
   x          pdf
## 0 4.734738e-01
## 1 3.818337e-01
## 2 1.231722e-01
## 3 1.986648e-02
## 4 1.602135e-03
## 5 5.168179e-05

$P(X=4)$ directly from your formula: (Do you have ${5 \choose 4} = 5?$)

p = 5/36;  choose(5,4)*p^4*(1-p)
## 0.001602135

Image from your Question: