Question 538727
Four dice are thrown. What is the probability that one of the rolls is a 6, given that one of the rolls is a 2?
<pre>

This is a conditional probability   P(A|B) = {{{"P(A_and_B)"/"P(B)"}}} = {{{"N(A_and_B)"/"N(B)"}}}



               number of rolls with both a 2 and a 6  
Probability = --------------------------------------------
                     number of rolls with a 6

We calculate the denominator first:

denominator = {{{(matrix(7,1,number,of,possible,rolls,with,a,6))}}} = {{{(matrix(4,1,number,of,possible,rolls))}}} - {{{(matrix(7,1,number,of,possible,rolls,without,a,6))}}}

denominator = 6×6×6×6 - 5×5×5×5 = 6<sup>4</sup> - 5<sup>4</sup> = 1296 - 625 = 671

numerator = {{{(matrix(11,1,number,of,possible,rolls,with,both,a,6,and, a,2))}}}

We use the formula 

     N(A or B) = N(A) + N(B) - N(A and B) 

Solving for N(A and B),

    N(A and B) = N(A) + N(B) - N(A or B)     

numerator = {{{(matrix(11,1,number,of,possible,rolls,with,both,a,2,and, a,6))}}} = {{{(matrix(7,1,number,of,possible,rolls,with,a,2))}}} + {{{(matrix(7,1,number,of,possible,rolls,with,a,6))}}} - {{{(matrix(10,1,number,of,possible,rolls,with,a,2,or,a,6))}}}

We have already calculated {{{(matrix(7,1,number,of,possible,rolls,with,a,6))}}} = 671

and

{{{(matrix(7,1,number,of,possible,rolls,with,a,2))}}} is the same quantity, that is, it is also 671

Now we only need to calculate {{{(matrix(10,1,number,of,possible,rolls,with,a,2,or,a,6))}}}

{{{(matrix(10,1,number,of,possible,rolls,with,a,2,or,a,6))}}} = {{{(matrix(4,1,number,of,possible,rolls))}}} - {{{(matrix(11,1,number,of,possible,rolls,with,neither,a,2,nor,a,6))}}} = 6<sup>4</sup> - 4<sup>4</sup> = 1296 - 256 = 1040

Now we go back to:

numerator = {{{(matrix(11,1,number,of,possible,rolls,with,both,a,2,and, a,6))}}} = {{{(matrix(7,1,number,of,possible,rolls,with,a,2))}}} + {{{(matrix(7,1,number,of,possible,rolls,with,a,6))}}} - {{{(matrix(10,1,number,of,possible,rolls,with,a,2,or,a,6))}}} = 671 + 671 - 1040 = 302.
 
Therefore the desired probability is {{{302/671}}}.

Edwin</pre>