Question

A dice is thrown 2times. What will be the probability of getting 11?

Answer 1

`"P"("11 in two rolls")=1/18`, or approx. 5.56%.

Explanation:

We assume the die has 6 sides, the die is fair, and the two rolls are independent.

In order to get a total of 11 in two rolls, we need one of two things to happen:

• the first roll is 6, and the second roll is 5; or
• the first roll is 5, and the second roll is 6.

The probability of the first throw being a 6 is `1/6,` and the probability of the second throw being a 5 is also `1/6.` Since the two throws are assumed independent, we get

`"P"("roll 6, then 5")="P"("roll a 6") xx "P"("roll a 5")`
`color(white)("P"("roll 6, then 5"))="       "1/6"       "xx"       "1/6`
`color(white)("P"("roll 6, then 5"))=1/36`

The probability of rolling a 5 first, then a 6, is calculated exactly the same way; it is also `1/36.`

So we get

`"P"("11 in two rolls")="P"("roll 6, then 5") + "P"("roll 5, then 6")`

`color(white)("P"("11 in two rolls"))=1/36 + 1/36`

`color(white)("P"("11 in two rolls"))=2/36"           " = 1/18`

This can be written as an (approximate) percentage as 5.56%.

Answer 2

An alternative approach:

Explanation:

An alternate way to find this is to chart the potential results:

`((color(white)(0),ul1,ul2,ul3,ul4,ul5,ul6),(1|,2,3,4,5,6,7),(2|,3,4,5,6,7,8),(3|,4,5,6,7,8,9),(4|,5,6,7,8,9,10),(5|,6,7,8,9,10,11),(6|,7,8,9,10,11,12))`

Notice that there are 2 ways to get a sum of eleven out of 36 possible rolls, or `2/36=1/18`