Question

When two dice are rolled, how do you find the probability of getting a sum less than 11?

Answer 1

`P("less than 11") = 33/36 = 11/12`

Explanation:

If 2 dice are thrown, there are `6xx6 = 36` outcomes.

There is only one way to get a total of 12.
There are only two ways to get a total of 11. `5+6 " or " 6+5`

Therefore of the 36 possible outcomes there are 3 that do not meet the requirement of being less than 11.

`P("less than 11") = 33/36 = 11/12`

However, for similar questions which might ask

`rarr` both are prime
`rarr` a prime and multiple of 3
`rarr` a prime and a square, etc etc

I like the method of using a "possibility space".
This is a diagram with two axes which shows the outcomes of the dice and the possible combinations. (hence "possibility" space.

In this way all the outcomes are shown.
The time taken to draw the space is made up by the ease wit which the answers can be found.

I have used red die and a blue die to illustrate

`color(red)(darr"red die")`
`color(red)(6)color(white)(xx x)7" "8" "9" "10" "11" "12`

`color(red)(5)color(white)(xx x)6" "7" "8" "9" "10" "11`

`color(red)(4)color(white)(xx x)5" "6" "7" "8" "9" "10`

`color(red)(3)color(white)(xx x)4 " "5" "6" "7" "8" "9`

`color(red)(2)color(white)(xx x)3" "4 " "5" "6" "7" "8`

`color(red)(1)color(white)(xx x)2" "3" "4 " "5" "6" "7`

`color(white)(xxxx)color(blue)(1" "2" "3 " "4" "5" "6 larr"blue die")`

The values in the grid represent the sum of the numbers on 2 dice.

Notice: there are `6xx6 = 36` outcomes.
There are 33 outcomes less than 11.