Math Is Fun Forum

14_karat

Member

Registered: 2013-10-01

Posts: 1

Probability

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!

Bob

Administrator

Registered: 2010-06-20

Posts: 10,619

Re: Probability

hi 14_karat

Stay on-line while I put together an explanation. 

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

Bob

Administrator

Registered: 2010-06-20

Posts: 10,619

Re: Probability

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

---

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

auyeungyat

Member

Registered: 2013-09-23

Posts: 22

Re: Probability

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)

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Probability

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.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

auyeungyat

Member

Registered: 2013-09-23

Posts: 22

Re: Probability

So,WHO IS RIGHT!!!

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Probability

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.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,619

Re: Probability

So,WHO IS RIGHT!!!

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

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