Question 3. Three unusual dice, A, B, and C are constructed such that die A has

Question

Question 3.

Three unusual dice, A, B, and C are constructed such that die A has the numbers 3, 3, 4, 4, 8, 8; die B has

the numbers 1, 1, 5, 5, 9, 9; and die C the numbers 3, 5, 6, 6, 7, 9.
(a) If dice A and B are rolled, find the probability that B beats A, that is, the number that appears on
die B is greater than the number on die A.
(b) If dice C and B are rolled, find the probability that the same number appears on both die B and C.
(c) If dice A and C are rolled, find the probability that the sum of the numbers is greater than 10.

Explanation / Answer

(a) If Dice A and B ae rolled,

Probability that B beats A, it will beat A when B has higher number than A .

Pr(B beats A) has sample space (3,5), (4,5), (8,9) ,(3,9), (4,9) WHere first number if on A and second number if on B.

Pr(B beats A) = Pr(3,5) + Pr(4,5) + Pr(8,9) + Pr(3,9) + Pr(4,9)

In short we can say that if there is number 9 on Dice B, then it is 100% probability that B will beat A, if number 5 on dice B, then if number 5 on dice B, then die must have either 3 or 4 than B will beat A.

Pr(B beats A) = Pr(B is 9) * Pr(ANy number on A) +Pr(B is 5) * Pr(A is 3 or 4)

= (1/3) * 1 + 1/3 * (2/3) = 1/3 + 2/9 = 5/9

(b) Here the sample space on B and C are (5,5) , (9,9)

Pr(Both Die B and C have same number) = Pr(5 on B) * Pr(5 on C) + Pr(9 on B) * Pr(9 on C)

= 2/6 * 1/6 + 2/6 * 1/6 = 4/36 = 1/9

(c) If dice A and C are rolled, Sum is greater than 10.

so sample space = (3,9),(4,7), (4,9) ,(8,3), (8,5),(8,6), (8,7) , (8,9)

so we can see that whenever there is 8 on A, the sum will be greater than 10, when there is 4 on dice A, Dice B must have 7 or 9 . when Dice A have 3 on it, DIce B should have 9.

Pr(Sum is greater than 10) = Pr(Dice A have 8) * Pr(Dice B have anything) + Pr(Dice A have 4) * Pr(DIce B have 7 and 9) + Pr(Dice A have 3) * Pr(DIce C have 3)

= 1/3 * 1 + 1/3 * 1/3 + 1/3 * 1/6 = 1/2 = 0.5

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