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) Here sum of the two rolls is greater than 9.

So sample space : (5,5), (4,6),(6,4), (5,6),(6,5) ,(6,6)

these are the possible combination of rolls when dice is rolled twice.

Pr(Sum > 9) = Pr(5,5) + Pr(4,6) + Pr(6,4) + Pr(5,6) + Pr(6,5) + Pr(6,6)

= 1/8 * 1/8 + 1/8 * 1/4 * 2 + 2 * 1/8 * 1/4 + 1/4 * 1/4 = 13/64

(b) Here maximum value on two rolls is less than 3 ; that means 1 or 2 would be the maximum value.

Sample space : (1,1) , (1,2) ,(2,1) , (2,2)

Pr(Max value < 3) = Pr(1,1) + Pr(1,2) + Pr(2,1) + Pr(2,2)

= 1/4 * 1/4 + 1/4 * 1/8 + 1/4 * 1/8 + 1/8 * 1/8 = 9/64

(c) Minimum value to appear on the two rolls is not less than 4; so that would be 5 or 6

sample space : (5,5) , (5,6), (6,5) , (6,6)

Pr(Minium value > 4) = Pr(5,5) + Pr(5,6) + Pr(6,5) + Pr(6,6)

= 1/8 * 1/8 + 1/8 * 1/4 + 1/4 * 1/8 + 1/4 * 1/4 = 9/64

(d) Here first roll shall be greater than the second roll.

Sampe spce : (2,1) , (3,2), (3,1) , (4,1) ,(4,2) ,(4,3), (5,1),(5,2) ,(5,3), (5,4) , (6,1) ,(6,2) ,(6,3) ,(6,4), (6,5)

Pr(First roll is greater than second roll) = Pr(2,1) + Pr(3,2) + Pr(3,1) + Pr(4,1) + Pr(4,2) + Pr((4,3) + Pr(5,1) + Pr(5,2) + Pr(5,3) + Pr (5,4) +Pr (6,1) + Pr(6,2) +Pr(6,3) +Pr(6,4) +Pr (6,5)

= 1/8 * 1/4 + 1/8 * (1/4 + 1/8) + 1/8 * (1/4 + 1/8 + 1/8) + 1/8 * (1/4 + 1/8 + 1/8 + 1/8) + 1/4 * (1/4 + 1/8 + 1/8 + 1/8 + 1/8) = 13/64

(e) First roll is equal to secon roll.

Sample space : (1,1) ,(2,2) , (3,3) , (4,4) , (5,5) , (6,6)

Pr(first roll = second roll) = Pr(1,1) + Pr(2,2) + Pr(3,3) + Pr(4,4) + Pr(5,5) + Pr(6,6)

= 1/4 * 1/4 + 1/8 * 1/8 + 1/8 * 1/8 + 1/8 * 1/8 + 1/8 * 1/8 + 1/4 * 1/4 = 3/16

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