Math Is Fun Forum

qDOS

Member

Registered: 2013-12-22

Posts: 2

Probability with multiple variables

Greetings,

I'm wondering how I might calculate the probability of rolling at least one '10' when rolling three ten-sided dice. I know it cant simply be (3x10)% chance since rolling ten dice wont give me a 100% chance of rolling at least one 10.

Can you help me? I'd like to understand how to do problems like this.

Happy holidays!

Last edited by qDOS (2013-12-22 21:35:43)

Bob

Administrator

Registered: 2010-06-20

Posts: 10,164

Re: Probability with multiple variables

hi qDOS

Welcome to the forum.

Here's a diagram which will show a way of doing these.

I'll edit in an explanation in the next few minutes.

Edit:

The diagram is called a tree diagram ( I guess because it has branches).

First decide how many events.  You want to 'throw' three dice so that's one set of branches for each die.

Next decide how many outcomes you're interested in for each die.  You just want to know: is it a ten or not, so that's a pair of branches for each die.

Assuming the dice have ten numbers and are not weighted in some biasing way then the probability of throwing a ten will be 1/10 and of not throwing a ten 9/10.

The second and third dice behave independently from the first so the probabilities will be the same for each stage.

To get the probability of travelling along any route of the diagram (left to right) you multiply the individual probabilities together.

So, for example, the probability of ten ten ten will be 1/10 x 1/10 x 1/10 = 1/1000 or 0.001 or 0.1%

If you work out all eight probabilities like this they will always add up to 1 (since it is certain that one of them will happen).

To answer your question, it is quickest to work out P(no tens) and subtract from 1.

Bob

Last edited by Bob (2013-12-22 22:55:19)

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

qDOS

Member

Registered: 2013-12-22

Posts: 2

Re: Probability with multiple variables

Hm, so P(no tens) would be (9/10)^3. And 1 - (9/10)^3 makes 27.1%

That seems reasonable I expected it to be something just short of 30%

Thanks a lot!

Bob

Administrator

Registered: 2010-06-20

Posts: 10,164

Re: Probability with multiple variables

hi qDOS,

That's correct!  Well done

This is the Maths Is Fun page on tree diagrams:

http://www.mathsisfun.com/data/probabil … grams.html

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob