Suppose a biased die could cause the occurrence of number 1 & 6 with higher prob

Question

Suppose a biased die could cause the occurrence of number 1 & 6 with higher probability The probability distribution is listed as below Number 1 6 Probability 48 8 84 If the biased die is rolled twice, what is the probability that: a) Sum of the two rolls is greater than 9; b) The maximum value to appear on the two rolls is less than 3 cThe minimum value to appear on the two rolls is not less than 4 d) The value of first roll is greater than the value of the second roll e) The value of first roll is equal to the value of the second roll

Explanation / Answer

(a) Sum of two rolls would be greater than 9 in the following cases: (4,6), (5,5), (5,6), (6,4), (6,5) and (6,6)  

The corresponding probabilities are: (1/8) x (1/4); (1/8) x (1/8); (1/8) x (1/4); (1/4) x (1/8); (1/4) x (1/8) and (1/4) x (1/4)

The required probability = 1/32 + 1/64 + 1/32 + 1/32 + 1/32 + 1/16 = (2+1+2+2+2+4)/64 = 13/64

(b) The maximum value appearing on the two rolls would be less than 3 in the following cases: (1,1), (1,2), (2,1), (2,2). We are considering the highest value of the two rolls to be less than 3, and not the sum of the values of the ywo rolls to be less than 3.

The required probability = (1/4) x (1/4) + (1/4) x (1/8) + (1/8) x (1/4) + (1/8) x (1/8) = (4+2+2+1)/64 = 9/64

(c) The minimum value to appear on the two rolls is not less than 4 in the following cases:

(4,4), (4,5), (4,6), (5,4), (5,5), (5,6), (6,4), (6,5), (6,6).

The required probability = (1/8) x (1/8) + (1/8) x (1/8) + (1/8) x (1/4) + (1/8) x (1/8) + (1/8) x (1/8) + (1/8) x (1/4) + (1/4) x (1/8) + (1/4) x (1/8) + (1/4) x (1/4) = (1+1+2+1+1+2+2+2+4)/64 = 16/64 = 1/4

(d) The value of the first roll is greater than the value of the second roll in the following cases:

(2,1), (3,1), (3,2), (4,1), (4,2), (4,3), (5,1), (5,2), (5,3), (5,4), (6,1), (6,2), (6,3), (6,4), (6,5)

The required probability = (1/8) x (1/4) + (1/8) x (1/4) + (1/8) x (1/8)+ (1/8) x (1/4) +(1/8) x (1/8) + (1/8) x (1/8) + (1/8) x (1/4) + (1/8) x (1/8) + (1/8) x (1/8) + (1/8) x (1/8) + (1/4) x (1/4) + (1/4) x (1/8) + (1/4) x (1/8) + (1/4) x (1/8) + (1/4) x (1/8) = (2+2+1+2+1+1+2+1+1+1+4+2+2+2+2)/64 = 26/64 = 13/32

(e) The value of the first roll is equal to the value of the second roll in the following cases:

(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)

The required probability = (1/4) x (1/4) + (1/8) x (1/8) + (1/8) x (1/8) + (1/8) x (1/8) + (1/8) x (1/8) + (1/4) x (1/4) = (4+1+1+1+1+4)/64 = 12/64 = 3/16

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