What is the probability that in three consecutive rolls of two fair dice, a person gets a total of 7, followed by a total of 11, followed by a total of 7?

Let us consider the first case. We denote the probability of "getting a total of 7 in a throw of 2 fair dices" by #P(A)#.

Now, there are 6 outcomes when we throw a dice, so the number of possible outcomes when we throw 2 dices at a time is

#6*6=6^2=36#

Now by observation, we get

#1+6=7, 2+5=7, 3+4=7 #

They can interchange their position in #2!# ways, so the number of all possible cases in favour of #A# is

#3*2=6#

Hence

#P(A) = 6/36 =1/6#

Let us consider the 2nd case. We denote the probability of "getting a total of 11 in a throw of 2 fair dices" by #P(B)#.

Now, there are 6 outcomes when we throw a dice, so the number of possible outcomes when we throw 2 dices at a time is

#6*6=6^2=36#

Now by observation, we get

#5+6=11#

They can interchange their position in #2!# ways, so the number of all possible cases in favour of #B# is

#1*2=2#

Hence

#P(B) = 2/36 =1/18#

Let us consider the 3rd case. It is the same as the first case, hence

#P(C) = 6/36 =1/6#

The required probability asked in the question is #P(A nn B nn C)#.

The events A, B, C are independent, so

#P(A nn B nn C) = P(A)*P(B)*P(C) = 1/6*1/18*1/6 = 1/648#

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

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

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

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

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

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

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

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

There are #6xx6= 36# possible outcomes

There are #6# ways of rolling #7# and #2# ways of rolling 11.

#P(7,11,7) = 6/36 xx 2/36 xx 6/36#

#= 1/6 xx1/18 xx1/6#

#= 1/648#