# If two dice are rolled once and one die rolled a second time if the first two rolls have same number, what is the probability that the two or three rolls will sum to over 15?

Oct 28, 2015

probability $= \frac{1}{18}$

#### Explanation:

If the first two rolls have the same value there are only 2 possibilities for which the sum of the 3 rolls exceeds 15.

{: (color(black)("roll 1"),color(black)("roll 2"),color(black)("roll 3")), (5, 5, 6), (6,6,6) :}

Given that the first two rolls are the same:
$\textcolor{w h i t e}{\text{XX")P("two 5's}} = \frac{1}{6}$
$\textcolor{w h i t e}{\text{XX")P("two 5's") & P("third roll a 6}} = \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}$

Similarly
$\textcolor{w h i t e}{\text{XX")P("two 6's") & P("third roll a 6}} = \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}$

Since these are disjoint occurrences
$\textcolor{w h i t e}{\text{XX")P("5-5-6" or "6-6-6}}$

$\textcolor{w h i t e}{\text{XXXXX")=P("5-5-6")+P("6-6-6}}$

$\textcolor{w h i t e}{\text{XXXXX}} = \frac{2}{36} = \frac{1}{18}$