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#1 2013-10-01 10:45:58

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!

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#2 2013-10-01 10:53:10

Bob
Administrator
Registered: 2010-06-20
Posts: 10,052

Re: Probability

hi 14_karat

Stay on-line while I put together an explanation.  smile

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 smile

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#3 2013-10-01 11:00:04

Bob
Administrator
Registered: 2010-06-20
Posts: 10,052

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 smile

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#4 2013-10-01 23:46:45

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)

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#5 2013-10-01 23:54:07

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.

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#6 2013-10-02 00:25:29

auyeungyat
Member
Registered: 2013-09-23
Posts: 22

Re: Probability

So,WHO IS RIGHT!!!

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#7 2013-10-02 00:29:20

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.

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#8 2013-10-02 02:07:41

Bob
Administrator
Registered: 2010-06-20
Posts: 10,052

Re: Probability

So,WHO IS RIGHT!!!

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

3yNxiFu.gif

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 smile

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