Question

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?

*Round to the nearest ten thousandth.

Answer 1

`1/6*1/18*1/6 = 1/648`

Explanation:

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`

Answer 2

`1/648`

Explanation:

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`