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Rolling a fair 6-sided die twice, what is the probability that the sum of the two rolls is exactly 6 given that the first one is an even number?

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  • $\begingroup$ @veereshpandey That seems implausible, seeing as $$P(X+Y=6|X=2)=\frac16$$ $$P(X+Y=6|X=4)=\frac16 $$ $$P(X+Y=6|X=6)=0$$ Looks like $P(X+Y=6|X\text{ is even})$ is between $0$ and $\frac16.$ My guess: $\frac19.$ $\endgroup$
    – bof
    May 5 '18 at 5:19
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First die can be 2, 4, or 6. Second any roll 1-6. So the sample space is $3 \cdot 6= 18$ Only two positive outcomes: {2,4} and {4,2}.

Probability $= \frac{2}{18} = \frac{1}{9}$

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Let

  • $A $: sum is $6$

  • $B $: first is even

It follows:

  • $ |A \cap B| = 2$

  • $ |B|= 3\cdot 6 = 18$ $$P(A|B) = \frac {|A\cap B|}{|B|}= 2/18 = 1/9$$

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The number of sample space for the event of rolling a fair 6-sided die twice is 6.6 = 36. Let

A = { the sum of the two rolls is exactly 6 }, {1,5}{2,4},{3,3},{4,2},{5,1}

B ={the first one is an even number} {2,:},{4,:},{6,:}

$A\cap B$ = {2,4}, {4,2}

P($A\cap B$) = 2/36.

P(B) = (6+6+6)/36

P(A|B) = P($A\cap B$)/P(B) = 2/18

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