SOLUTION: 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?

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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?
Answer by AnlytcPhil(1810) About Me  (Show Source):
You can put this solution on YOUR website!
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?

This is a conditional probability   P(A|B) = %22P%28A_and_B%29%22%2F%22P%28B%29%22 = %22N%28A_and_B%29%22%2F%22N%28B%29%22



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

We calculate the denominator first:

denominator = %28matrix%287%2C1%2Cnumber%2Cof%2Cpossible%2Crolls%2Cwith%2Ca%2C6%29%29 = %28matrix%284%2C1%2Cnumber%2Cof%2Cpossible%2Crolls%29%29 - %28matrix%287%2C1%2Cnumber%2Cof%2Cpossible%2Crolls%2Cwithout%2Ca%2C6%29%29

denominator = 6×6×6×6 - 5×5×5×5 = 64 - 54 = 1296 - 625 = 671

numerator = 

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 =  = %28matrix%287%2C1%2Cnumber%2Cof%2Cpossible%2Crolls%2Cwith%2Ca%2C2%29%29 + %28matrix%287%2C1%2Cnumber%2Cof%2Cpossible%2Crolls%2Cwith%2Ca%2C6%29%29 - %28matrix%2810%2C1%2Cnumber%2Cof%2Cpossible%2Crolls%2Cwith%2Ca%2C2%2Cor%2Ca%2C6%29%29

We have already calculated %28matrix%287%2C1%2Cnumber%2Cof%2Cpossible%2Crolls%2Cwith%2Ca%2C6%29%29 = 671

and

%28matrix%287%2C1%2Cnumber%2Cof%2Cpossible%2Crolls%2Cwith%2Ca%2C2%29%29 is the same quantity, that is, it is also 671

Now we only need to calculate %28matrix%2810%2C1%2Cnumber%2Cof%2Cpossible%2Crolls%2Cwith%2Ca%2C2%2Cor%2Ca%2C6%29%29

%28matrix%2810%2C1%2Cnumber%2Cof%2Cpossible%2Crolls%2Cwith%2Ca%2C2%2Cor%2Ca%2C6%29%29 = %28matrix%284%2C1%2Cnumber%2Cof%2Cpossible%2Crolls%29%29 -  = 64 - 44 = 1296 - 256 = 1040

Now we go back to:

numerator =  = %28matrix%287%2C1%2Cnumber%2Cof%2Cpossible%2Crolls%2Cwith%2Ca%2C2%29%29 + %28matrix%287%2C1%2Cnumber%2Cof%2Cpossible%2Crolls%2Cwith%2Ca%2C6%29%29 - %28matrix%2810%2C1%2Cnumber%2Cof%2Cpossible%2Crolls%2Cwith%2Ca%2C2%2Cor%2Ca%2C6%29%29 = 671 + 671 - 1040 = 302.
 
Therefore the desired probability is 302%2F671.

Edwin