If two dice rolled and the values are added, how do you arrange the following probabilities in ascending order? The sum is odd, the sum is not #7#, the sum is exactly #12#, the sum is less than #5#.

1 Answer
Aug 5, 2017

#P("less than 5") = 6/36 = 1/6#
#P("sum is odd") = 1/2#
#P("not 7") =30/36 =5/6#
#P("not 12") = 35/36#

Explanation:

A possibility space is a good way of showing the possible outcomes when two dice are rolled:

This allows you to count all the required outcomes.

#color(white)(........)ul(1" "2" "3" "4" "5" ")6#

#1:color(white)(.....)2" "3" "4" "5" "6" "7#

#2:color(white)(.....)3" "4" "5" "6" "7" "8#

#3:color(white)(.....)4" "5" "6" "7" "8" "9#

#4:color(white)(.....)5" "6" "7" "8" "9" "10#

#5:color(white)(.....)6" "7" "8" "9" "10" "11#

#6:color(white)(.....)7" "8" "9" "10" "11" "12#

There are #6xx6= 36# possible outcomes

#P("less than 5") = 6/36 = 1/6#

#P("sum is odd") = 1/2#

There are #6# ways of getting a sum of #7#,
so there are #36-6 = 30# ways of NOT getting #7#

#P("not 7") =30/36 =5/6#

There is only one way of getting #12#,
so #36-1 =35# ways of NOT getting #12#

#P("not 12") = 35/36#

So the required order is:

#P("less than 5") = 6/36 = 1/6#
#P("sum is odd") = 1/2#
#P("not 7") =30/36 = 5/6#
#P("not 12") = 35/36#