SOLUTION: George plays a game using a biased die which is twice as likely to land on an even number as on an odd number. The probabilities for the three even numbers are all equal and the

Question 1176740: George plays a game using a biased die which is twice as likely to land on
an even number as on an odd number. The probabilities for the three even
numbers are all equal and the probabilities for the three odd numbers are
all equal.
George throws the die once and calculates his score by the following method.
• If the number lands on is 3 or less, he multiplies the number by 3 and
adds 1.
• If the number lands on is more than 3, he multiplies the number by 2 and
subtracts 4.
George throws the die twice.
(i) Find the probability that the total of the scores on the two throws is
16.
(ii) Given that the total of the scores on the two throws is 16, find the
probability that the score on the first throw was 6.
Answer by Edwin McCravy(20063) (Show Source): You can put this solution on YOUR website!
George plays a game using a biased die which is twice as likely to land on
an even number as on an odd number. The probabilities for the three even
numbers are all equal and the probabilities for the three odd numbers are
all equal.

P(1)+P(2)+P(3)+P(4)+P(5)+P(6) = 1

Let the probability of an odd number be x.
Then the probability of an even number will be 2x.

x+2x+x+2x+x+2x = 1
            9x = 1
             x = 1/9

So P(odd) = 1/9 and P(even) = 2/9

George throws the die once and calculates his score by the following method.
• If the number lands on is 3 or less, he multiplies the number by 3 and
adds 1.
• If the number lands on is more than 3, he multiplies the number by 2 and
subtracts 4.
George throws the die twice.
(i) Find the probability that the total of the scores on the two throws is
16.

 

roll  P(roll)  Arithmetic  Score   
 1      1/9      1x3+1       4
 2      2/9      2x3+1       7
 3      1/9      3x3+1      10 
 4      2/9      4x2-4       4
 5      1/9      5x2-4       6
 6      2/9      6x2-4       8 

The only way is for the rolls to be (10 and 6), (6 and 10) or (8 and 8) 
   "S1" = "score on first roll"
"S2" = "score on second roll"

P(S1+S2=16) = P[(S1=10 AND S2=6) OR P(S1=6 AND S2=10) OR P(S1=8 and S2=8)] =
P(3)∙P(5) + P(5)∙P(3) + P(6)P(6) = (1/9)(1/9) + (1/9)(1/9) + (2/9)(2/9) =
1/81+1/81+4/81) = 6/81 which reduces to 2/27.

------------------------------

(ii) Given that the total of the scores on the two throws is 16, find the
probability that the score on the first throw was 6.


P(A|B) = P(A and B)/P(B)


P[(S1=10 AND S2=6) OR P(S1=6 AND S2=10) OR P(S1=8 AND S2=8) | P(S1=10)] =

(2/27)/(1/9) = (2/27)(9/1) = 18/27 = 2/3

Edwin

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