## Question

secondary 4 | A Maths

Anyone can contribute an answer, even non-tutors.

##### Sonia

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Date Posted: 3 months ago
Views: 29
Eric Nicholas K
3 months ago
Hi Sonia!

For this question, you must consider the possible outcomes. Max has summed up all the answers accurately.

We assume that the three dice are fair (that is, nothing weird or unusual with the dice) and each die does not affect the other two dice.

There are a total of 6 x 6 x 6 = 216 possible outcomes of throws altogether. The possible outcomes are as follows.

First 1, second 1, third 1
First 1, second 1, third 2
First 1, second 1, third 3
First 1, second 1, third 4
First 1, second 1, third 5
First 1, second 1, third 6
First 1, second 2, third 1
First 1, second 2, third 2 and so on

...

First 6, second 6, third 5
First 6, second 6, third 6

Now, with this understanding, we proceed to answer the three parts.

(a) There is only one way in which the sum of the three numbers on the dice is 18.

"First 6, second 6, third 6"

This is one outcome out of 216 possible outcomes. Hence, the probability that the total sum is 18 is given by P (total sum is 18) = 1/216.

(b) There are six ways in which all three dice show the same number.

"First 1, second 1, third 1"
"First 2, second 2, third 2"
"First 3, second 3, third 3"
"First 4, second 4, third 4"
"First 5, second 5, third 5"
"First 6, second 6, third 6"

These account for six out of 216 possible outcomes. Hence, the probability that the three dice show the same number is given by P (all same number) = 6/216 which simplifies to 1/36.

(c) There are three ways in which the total sum is 17.

"First 5, second 6, third 6"
"First 6, second 5, third 6"
"First 6, second 6, third 5"

These account for three out of 216 possible outcomes. Hence, the probability that the three dice show the same number is given by P (total sum is 17) = 3/216 which simplifies to 1/72.

I hope the explanations are okay!