Question

In: Statistics and Probability

Three fair, six-sided dice colored red, green and blue are rolled. Calculate each of the following...

Three fair, six-sided dice colored red, green and blue are rolled. Calculate each of the following probabilities: (a) The probability all three dice show the same face (“triples”). (b) The probability that the red die shows a larger number than the green die. (c) The probability that the red die shows a larger number than the green die and the green die shows a larger number than the blue die. (d) The probability that the sum of the pips on all three dice is exactly 10. (e) The probability that the sum of the pips on all three dice is less than 10. (f) The probability that the sum of the pips on all three dice is greater than 10.

Expert Solution

There are 6 outcomes for the red die, 6 outcomes for the green die and 6 outcomes for the blue die. Hence by the Multiplication Principle there are 6×6×6=216 outcomes of the experiment.

Below is the sample space of all 216 outcomes.

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

Note: To understand lets take outcome (6,3,2) , which means red die shows 6 , green die shows 3 and blue die shows 2.

a) There are 6 possibliries that dice show same face ("triples") that are (1,1,1), (2,2,2), (3,3,3), (4,4,4), (5,5,5), (6,6,6).

The probability all three dice show the same face (“triples”) = = =

b) There are 6+12+18+24+30=90 possible outcomes in which red die shows a larger number than the green die, that are

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

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

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

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

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

The probability that the red die shows a larger number than the green die = = =

c) There are 1+3+6+10=20 possible outcomes in which the red die shows a larger number than the green die and the green die shows a larger number than the blue die, that are

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

The probability that the red die shows a larger number than the green die and the green die shows a larger number than the blue die = = =

d) There are 4+5+6+5+4+3 = 27 possible outcomes in which the sum of the pips on all three dice is exactly 10, that are

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

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

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

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

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

(6,1,3) (6,2,2) (6,3,1)

The probability that the sum of the pips on all three dice is exactly 10 = = =

e) There are 26+21+15+10+6+3 = 81 possible outcomes in which the sum of the pips on all three dice are less than 10, that are

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

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

The probability that the sum of the pips on all three dice is less than 10 =   =  =

f) There are 6+10+15+21+26+30 = 108 possible outcomes in which the sum of the pips on all three dice is exactly 10, that are

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

The probability that the sum of the pips on all three dice is greater than 10 =

  =   =

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