You roll 2 dice. What is the probability of getting a 3 or 6 on the second die, given that you rolled a 1 on the first die?

Question

You roll 2 dice.  What is the probability of getting a 3 or 6 on the second die, given that you rolled a 1 on the first die?

anonymous · Answer

what is the formula that you are using?  I am so confused on this!

amistre64 · Answer

1 2 3 4 5 6
given -> 1       x       x
              2
              3
              4
              5
              6

since we are restricting ourselves to a "1"; there are only 6 options to choose from in that category; and a 3 or 6 are 2 of them sooo...

2/6 = 1/3

amistre64 · Answer

the fancy stuff might be:$$P("3\ or\ 6"|1)\frac{P("3\ or\ 6"\ and\ 1)}{P(1)}$$

amistre64 · Answer

i think my chart is more intuitive tho

anonymous · Answer

I am still a bit confused.  If I have 12 chances on the first roll to get a 1, then I only have 2 chances to get a 3 or a 6, how does that really equate to 1/3.  I'm not seeing it.

amistre64 · Answer

12 chances?  regardless, it says, given that you roll a 1, so that part is taken care of; now we focus within the set of (1,x), which there are only 6 ways total

anonymous · Answer

so I don't have to worry about the probability of rolling a 1?  It's already figured out, so I just focus on the second part?
I only see 2 possibilities for the second roll, either a 3 or a 6.

amistre64 · Answer

the P(1) = 1/6

the P(3 or 6) = 2/6

the P("3 or 6" and 1) = 2/6 * 1/6 = 2/36

the P("3 or 6" | 1) = P("3 or 6" and 1) / P(1)

2/36     2/6
---- = ---- = 2/6 = 1/3
1/ 6       1

amistre64 · Answer

there are 36 total outcomes for this scenario:|dw:1315340263708:dw|