if two fair dice (with faces numbered 1,2,3,4,5,6) are tossed together, what is the probability that the total score will be (a) a prime number (b) a perfect square (c) a perfect cubc?
richmorichmo
17 Oct, 2019
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The odds of rolling one five from two dice rolls is 136. The odds of rolling an odd number from the sum of two rolls requires that we roll one even number from one die and an odd number from another die. The odds of this happening are 12.
The probability of showing a prime number on a single die is 1/2, hence the probability of not showing a prime number is also 1/2. Total possible outcomes when two dice are rolled = 6*6 = 36.
Perfect squares in a dice of 6 faces from 1–6 is 1 and 4. So Probability that you get a 1 or a 4 in any one dice simply means you have to find out all possible combinations and see how many times 1 and 4 appear. ... So probability you will get a 1 or 4 or both is 1 - (4/6)² = 5/9.
The smallest cubes are 1^3=1, 2^3=8, and 3^=27. With any dice, the minimum is two. Even with dodecahedron dice, the maximum is 24. This only leaves 8.
Sample space:
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) |(2) (3) (4) (5) (6) (7)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6) |(3) (4) (5) (6) (7) (8)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6) |(4) (5) (6) (7) (8) (9)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6) |(5) (6) (7) (8) (9) (10)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6) |(6) (7) (8) (9) (10) (11)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6) |(7) (8) (9) (10) (11) (12)
№ of possible outcome= № of sample space= n(s)= 36
(a) Probability(a prime number)= № of required prime numbers/№ of sample space= 14/36= 7/18
(b) Probability(a perfect square)= № of required perfect squares/№ of sample space= 7/36
(c) Probability(a perfect cube)= № of required perfect cubes/№ of sample space= 5/36
Thank you.