Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**14_karat****Member**- Registered: 2013-10-01
- Posts: 1

Three fair dice are tossed. Find the probability of the dice showing:

a. At most one 6

B. At least two 6's

Can you also explain how to do it?!

Thank you!

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,426

hi 14_karat

Stay on-line while I put together an explanation.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,426

OK, here we go. Firstly, welcome to the forum!

Now when you throw a fair die, P(six) = 1/6 and P(not a six) = 5/6

To get "at most one six" you need to consider 4 cases:

P(six, no six, no six) = 1/6 x 5/6 x 5/6

no six, six, no six

no six, no six, six

no six, no six, no six

I've shown one case in full. You need to complete the other cases, get the four probabilities, and add them up.

Part B is now easy because if you don't get "at most one six" then you must get "at least two sixes". So answer B = 1 minus answer A.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**auyeungyat****Member**- Registered: 2013-09-23
- Posts: 15

It is wrong because

1.P=[1/6(six)x5/6(not six)x5/6(not six)=**25/216**]

According to your message,2 should be P=[1-[answer 1]] ,so your answer is P=[1-25/216=**191/216**].

Can you think that **191/216** is right?

My answer is:

A:P=[1/6(six)x5/6(not six)x5/6(not six)=**25/216**]

B:P=[1/6(six)x1/6(six)x1(no six=5/6,six=1/6(*It can be a six because the question b need two sixes,so three is acceptable*)5/6+1/6=1)=**1/36**]

*Last edited by auyeungyat (2013-10-01 23:49:40)*

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,255

Hi;

I am getting:

P(at most one 6) = 25 / 27

P(at least 2 sixes) = 2 / 27

done by the binomial distribution.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

**Online**

**auyeungyat****Member**- Registered: 2013-09-23
- Posts: 15

So,WHO IS RIGHT!!!

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,255

Hi;

Here is another way.

Here is by direct count:

There are 16 with 2 or more sixes, so it is 16 / 216 = 2 / 27

{{1, 1, 1}, {1, 1, 2}, {1, 1, 3}, {1, 1, 4}, {1, 1, 5}, {1, 1, 6}, {1,

2, 1}, {1, 2, 2}, {1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 2, 6}, {1,

3, 1}, {1, 3, 2}, {1, 3, 3}, {1, 3, 4}, {1, 3, 5}, {1, 3, 6}, {1, 4,

1}, {1, 4, 2}, {1, 4, 3}, {1, 4, 4}, {1, 4, 5}, {1, 4, 6}, {1, 5,

1}, {1, 5, 2}, {1, 5, 3}, {1, 5, 4}, {1, 5, 5}, {1, 5, 6}, {1, 6,

1}, {1, 6, 2}, {1, 6, 3}, {1, 6, 4}, {1, 6, 5}, {2, 1, 1}, {2, 1,

2}, {2, 1, 3}, {2, 1, 4}, {2, 1, 5}, {2, 1, 6}, {2, 2, 1}, {2, 2,

2}, {2, 2, 3}, {2, 2, 4}, {2, 2, 5}, {2, 2, 6}, {2, 3, 1}, {2, 3,

2}, {2, 3, 3}, {2, 3, 4}, {2, 3, 5}, {2, 3, 6}, {2, 4, 1}, {2, 4,

2}, {2, 4, 3}, {2, 4, 4}, {2, 4, 5}, {2, 4, 6}, {2, 5, 1}, {2, 5,

2}, {2, 5, 3}, {2, 5, 4}, {2, 5, 5}, {2, 5, 6}, {2, 6, 1}, {2, 6,

2}, {2, 6, 3}, {2, 6, 4}, {2, 6, 5}, {3, 1, 1}, {3, 1, 2}, {3, 1,

3}, {3, 1, 4}, {3, 1, 5}, {3, 1, 6}, {3, 2, 1}, {3, 2, 2}, {3, 2,

3}, {3, 2, 4}, {3, 2, 5}, {3, 2, 6}, {3, 3, 1}, {3, 3, 2}, {3, 3,

3}, {3, 3, 4}, {3, 3, 5}, {3, 3, 6}, {3, 4, 1}, {3, 4, 2}, {3, 4,

3}, {3, 4, 4}, {3, 4, 5}, {3, 4, 6}, {3, 5, 1}, {3, 5, 2}, {3, 5,

3}, {3, 5, 4}, {3, 5, 5}, {3, 5, 6}, {3, 6, 1}, {3, 6, 2}, {3, 6,

3}, {3, 6, 4}, {3, 6, 5}, {4, 1, 1}, {4, 1, 2}, {4, 1, 3}, {4, 1,

4}, {4, 1, 5}, {4, 1, 6}, {4, 2, 1}, {4, 2, 2}, {4, 2, 3}, {4, 2,

4}, {4, 2, 5}, {4, 2, 6}, {4, 3, 1}, {4, 3, 2}, {4, 3, 3}, {4, 3,

4}, {4, 3, 5}, {4, 3, 6}, {4, 4, 1}, {4, 4, 2}, {4, 4, 3}, {4, 4,

4}, {4, 4, 5}, {4, 4, 6}, {4, 5, 1}, {4, 5, 2}, {4, 5, 3}, {4, 5,

4}, {4, 5, 5}, {4, 5, 6}, {4, 6, 1}, {4, 6, 2}, {4, 6, 3}, {4, 6,

4}, {4, 6, 5}, {5, 1, 1}, {5, 1, 2}, {5, 1, 3}, {5, 1, 4}, {5, 1,

5}, {5, 1, 6}, {5, 2, 1}, {5, 2, 2}, {5, 2, 3}, {5, 2, 4}, {5, 2,

5}, {5, 2, 6}, {5, 3, 1}, {5, 3, 2}, {5, 3, 3}, {5, 3, 4}, {5, 3,

5}, {5, 3, 6}, {5, 4, 1}, {5, 4, 2}, {5, 4, 3}, {5, 4, 4}, {5, 4,

5}, {5, 4, 6}, {5, 5, 1}, {5, 5, 2}, {5, 5, 3}, {5, 5, 4}, {5, 5,

5}, {5, 5, 6}, {5, 6, 1}, {5, 6, 2}, {5, 6, 3}, {5, 6, 4}, {5, 6,

5}, {6, 1, 1}, {6, 1, 2}, {6, 1, 3}, {6, 1, 4}, {6, 1, 5}, {6, 2,

1}, {6, 2, 2}, {6, 2, 3}, {6, 2, 4}, {6, 2, 5}, {6, 3, 1}, {6, 3,

2}, {6, 3, 3}, {6, 3, 4}, {6, 3, 5}, {6, 4, 1}, {6, 4, 2}, {6, 4,

3}, {6, 4, 4}, {6, 4, 5}, {6, 5, 1}, {6, 5, 2}, {6, 5, 3}, {6, 5,

4}, {6, 5, 5}}

There are 200 having one or no sixes. 200 / 216 = 25 / 27

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

**Online**

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,426

So,WHO IS RIGHT!!!

Well both of course! as they lead to the same answer.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

Pages: **1**