What is the probability of getting a sum of either 7, 11, or 12 on a roll of two dice?

2 Answers
Feb 4, 2016

Answer:

The probability is #25%#.

Explanation:

Let's first take a look at the probability for one of those sums.

There are #6 times 6 = 36# different results of a roll of two dice:

#(1,1), (1,2), ..., (1, 6)#
#(2, 1), (2, 2), ..., (2,6)#
#...#
#(6,1), (6,2), ..., (6,6)#

The probability of each one of those is #1/36#.

  • How many possible combinations of two dice will give you a sum of #7#? There are #6# combinations: #(1,6)#, #(6,1)#, #(2,5)#, #(5,2)#, # (3,4)# and #(4,3)#.

#=> P("sum"=7) = 6 * 1/36 = 6/36 = 1/6#

  • For a sum of #11#, there are #2# combinations: #(5,6)# and #(6,5)#.

#=> P("sum"=11) = 2 * 1/36 = 2/36 = 1/18#

  • For a sum of #12#, there is just #1# combinations: #(6,6)#.

#=> P("sum"=12) = 1/36#

Now, how do you combine those three probabilities?

The events "#"sum"=7#", "#"sum"=11#" and "#"sum"=12#" are independent events since neither of them can ever occur at the same time.

For independent events #A# and #B# it holds

#P(A " or " B) = P(A) + P(B)#

Thus, our probability is

#P = P("sum"=7) + P("sum"=11) + P("sum"=12)#

#= 6/36 + 2/36 + 1/36 = 9/36#

#= 1/4#

#= 25%#

May 13, 2018

Answer:

#P(7,11,12) = 9/36 = 1/4#

Explanation:

You can draw up a possibility space and count how many of the outcomes meet the requirements.

The following array shows the sum of two dice thrown.
(Created by Parzival)

#{: (color(white)(0)," "ul1" "," "ul2" "," "ul3" "," "ul4" "," "ul5" "," "ul6" "),(1|," "2" "," "3" "," "4" "," "5" "," "6" "," "color(red)(7)" "),(2|," "3" "," "4" "," "5" "," "6" "," "color(red)(7)" "," "8" "),(3|," "4" "," "5" "," "6" "," "color(red)(7)" "," "8" "," "9" "),(4|," "5" "," "6" "," "color(red)(7)" "," "8" "," "9" "," "10" "),(5|," "6" "," "color(red)(7)" "," "8" "," "9" "," "10" "," "color(red)(11)" "),(6|," "color(red)(7)" "," "8" "," "9" "," "10" "," "color(red)(11)" "," "color(red)(12)" ") :}#

There are #6xx6 =36# outcomes

Of these there are #9# which add to #7, 11 or 12#

#P(7,11,12) = 9/36 = 1/4#