Question

1

Answer 1

these sound like conditional probability problems.

let's look at the first one:

if the other two dice ar fair, then the probability of rolling a 6 is 1/6, right?

so what we can do is condition...

\[P( \text{ rolling two }6\text{'s}) \]\[= P(\text{ rolling two }6\text{'s}\cap \text{ fair die is selected})+P(\text{ rolling two }6\text{'s}\cap \text{ weighted die is selected})\]\[=P(\text{ rolling two }6\text{'s}| \text{ fair die is selected})\cdot P( \text{ fair die is selected}) \]\[+ P(\text{ rolling two }6\text{'s}| \text{ weighted die is selected}) \cdot P( \text{ weighted die is selected}) \]
\[=\left( \frac{ 1 }{ 36 } \right)\cdot \frac{ 2 }{ 3 }+\left( \frac{ 1 }{ 64 } \right)\cdot \frac{ 1 }{ 3 }\]

Answer 2

for part b, we're going to need to compute this:
\[P \left( \text{ rolled the weighted die}|\text{ rolled two }6\text{'s} \right)=\frac{ P \left( \text{ rolled the weighted die}\cap\text{ rolled two }6\text{'s} \right) }{ P \left( \text{ rolled two }6\text{'s} \right) }\]

as you can see, the denominator in this one is the answer from part a, and the numerator is just 1/64.

I hope this makes sense.

Answer 3

@0202 please have a look...

Answer 4

@pgpilot326 omg yes it finally makes sense! thanks so much...i'm gonna try to do it myself now

Answer 5

cool! do you understand the partitioning...

breaking up rolling two 6's into the sum of the intersections?