What is the probability of rolling a number greater than or equal to 8 with the sum of two dice, given that at least one of the dice must show a 6?

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Dec 18, 2017

Answer:

$\frac{9}{11} \left(A n s w e r D .\right)$

Explanation:

$P \left[\text{sum" >= 8|"at least one dice shows 6}\right]$
$= \text{P[sum >= 8 AND at least one dice shows 6]" / "P[at least one dice shows 6]}$
$= \frac{\frac{9}{36}}{1 - {\left(\frac{5}{6}\right)}^{2}}$
$= \frac{\frac{1}{4}}{\frac{11}{36}}$
$= \frac{9}{11}$
$\left(A n s w e r D .\right)$

$P \left[\text{sum>=8 AND at least one dice shows 6}\right] =$
$\text{9/36 because there are 9 good combinations out of the 36 :}$
$\left(6 , 2\right) , \left(2 , 6\right) , \left(6 , 3\right) , \left(3 , 6\right) , \left(6 , 4\right) , \left(4 , 6\right) , \left(6 , 5\right) , \left(5 , 6\right) , \mathmr{and} \left(6 , 6\right) .$

$P \left[\text{at least one dice shows 6"] = 1 - P["no dice shows 6}\right]$
$= 1 - {\left(\frac{5}{6}\right)}^{2}$
$\text{Another possibility is using the formula for OR :}$
$\text{P[dice 1 is 6 OR dice 2 is 6]}$
$\text{= P[dice 1 is 6] + P[dice 2 is 6] - P[ dice 1 AND dice 2 is 6]}$
$= \frac{1}{6} + \frac{1}{6} - \frac{1}{6 \cdot 6}$
$= \frac{6 + 6 - 1}{36}$
$= \frac{11}{36}$

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