# A pair of fair six-sided dice is thrown eight times. Find the probability that a score greater than 7 is scored no more than five times?

$\cong 0.9391$

#### Explanation:

Before we get into the question itself, let's talk about the method for solving it.

Let's say, for instance, that I want to account for all the possible results from flipping a fair coin three times. I can get HHH, TTT, TTH, and HHT.

The probability of H is $\frac{1}{2}$ and the probability for T is also $\frac{1}{2}$.

For HHH and for TTT, that is $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$ each.

For TTH and HHT, it's also $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$ each, but since there are 3 ways I can get each result, it ends up being $3 \times \frac{1}{8} = \frac{3}{8}$ each.

When I sum up these results, I get $\frac{1}{8} + \frac{3}{8} + \frac{3}{8} + \frac{1}{8} = 1$ - which means I now have all the possible results of the coin flip accounted for.

Notice that if I set $H$ to be $p$ and therefore have $T$ be ~p, and also notice that we have a line from the Pascal's Triangle $\left(1 , 3 , 3 , 1\right)$, we've set up a form of:

sum_(k=0)^(n)C_(n,k)(p)^k((~p)^(n-k))

and so in this example, we get:

$= {C}_{3 , 0} {\left(\frac{1}{2}\right)}^{0} {\left(\frac{1}{2}\right)}^{3} + {C}_{3 , 1} {\left(\frac{1}{2}\right)}^{1} {\left(\frac{1}{2}\right)}^{2} + {C}_{3 , 2} {\left(\frac{1}{2}\right)}^{2} {\left(\frac{1}{2}\right)}^{1} + {C}_{3 , 3} {\left(\frac{1}{2}\right)}^{3} {\left(\frac{1}{2}\right)}^{0}$

$= 1 \left(1\right) \left(\frac{1}{8}\right) + 3 \left(\frac{1}{2}\right) \left(\frac{1}{4}\right) + 3 \left(\frac{1}{4}\right) \left(\frac{1}{2}\right) + 1 \left(\frac{1}{8}\right) \left(1\right)$

$= \frac{1}{8} + \frac{3}{8} + \frac{3}{8} + \frac{1}{8} = 1$

Now we can do the problem.

We're given the number of rolls as 8, so $n = 8$.

$p$ is the sum greater than 7. To find the probability of getting a sum greater than 7, let's look at the possible rolls:

$\left(\begin{matrix}\textcolor{w h i t e}{0} & \underline{1} & \underline{2} & \underline{3} & \underline{4} & \underline{5} & \underline{6} \\ 1 | & 2 & 3 & 4 & 5 & 6 & 7 \\ 2 | & 3 & 4 & 5 & 6 & 7 & 8 \\ 3 | & 4 & 5 & 6 & 7 & 8 & 9 \\ 4 | & 5 & 6 & 7 & 8 & 9 & 10 \\ 5 | & 6 & 7 & 8 & 9 & 10 & 11 \\ 6 | & 7 & 8 & 9 & 10 & 11 & 12\end{matrix}\right)$

Out of 36 possibilities, 15 rolls give a sum greater than 36, giving a probability of $\frac{15}{36} = \frac{5}{12}$.

With p=5/12, ~p=7/12

We can write out the entire sum of possibilities - from getting all 8 rolls being a sum greater than 7 all the way to getting all 8 rolls being a sum of 7 or less:

$= {C}_{8 , 0} {\left(\frac{5}{12}\right)}^{8} {\left(\frac{7}{12}\right)}^{0} + {C}_{8 , 1} {\left(\frac{5}{12}\right)}^{7} {\left(\frac{7}{12}\right)}^{1} + {C}_{8 , 2} {\left(\frac{5}{12}\right)}^{6} {\left(\frac{7}{12}\right)}^{2} + {C}_{8 , 3} {\left(\frac{5}{12}\right)}^{5} {\left(\frac{7}{12}\right)}^{3} + {C}_{8 , 4} {\left(\frac{5}{12}\right)}^{4} {\left(\frac{7}{12}\right)}^{4} + {C}_{8 , 5} {\left(\frac{5}{12}\right)}^{3} {\left(\frac{7}{12}\right)}^{5} + {C}_{8 , 6} {\left(\frac{5}{12}\right)}^{2} {\left(\frac{7}{12}\right)}^{6} + {C}_{8 , 7} {\left(\frac{5}{12}\right)}^{1} {\left(\frac{7}{12}\right)}^{7} + {C}_{8 , 8} {\left(\frac{5}{12}\right)}^{0} {\left(\frac{7}{12}\right)}^{8} = 1$

but we're interested in summing up only those terms that have our greater than 7 sum happening 5 times or less:

$= {C}_{8 , 3} {\left(\frac{5}{12}\right)}^{5} {\left(\frac{7}{12}\right)}^{3} + {C}_{8 , 4} {\left(\frac{5}{12}\right)}^{4} {\left(\frac{7}{12}\right)}^{4} + {C}_{8 , 5} {\left(\frac{5}{12}\right)}^{3} {\left(\frac{7}{12}\right)}^{5} + {C}_{8 , 6} {\left(\frac{5}{12}\right)}^{2} {\left(\frac{7}{12}\right)}^{6} + {C}_{8 , 7} {\left(\frac{5}{12}\right)}^{1} {\left(\frac{7}{12}\right)}^{7} + {C}_{8 , 8} {\left(\frac{5}{12}\right)}^{0} {\left(\frac{7}{12}\right)}^{8}$

$\cong 0.9391$

Feb 10, 2018

$0.93906$

#### Explanation:

$\text{So P[ outcome > 7 ]= 15/36 = 5/12}$
P["it occurs k times on 8 throws"] = C(8,k) (5/12)^k (7/12)^(8-k)"
$\text{(binomial distribution)}$
$\text{with "C(n,k) = (n!)/((n-k)! k!) " (combinations)}$
$\text{So, }$
$P \left[\text{it occurs at most 5 times on 8 throws}\right]$
$= 1 - P \left[\text{it occurs 6, 7, or 8 times on 8 throws}\right]$
$= 1 - C \left(8 , 6\right) {\left(\frac{5}{12}\right)}^{6} {\left(\frac{7}{12}\right)}^{2} - C \left(8 , 7\right) {\left(\frac{5}{12}\right)}^{7} \left(\frac{7}{12}\right) - {\left(\frac{5}{12}\right)}^{8}$
$= 1 - {\left(\frac{5}{12}\right)}^{8} \left(1 + 8 \cdot \left(\frac{7}{5}\right) + 28 \cdot {\left(\frac{7}{5}\right)}^{2}\right)$
$= 1 - {\left(\frac{5}{12}\right)}^{8} \left(1 + 11.2 + 54.88\right) = 1 - {\left(\frac{5}{12}\right)}^{8} \left(67.08\right)$
$= 0.93906$