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Topic review (newest first)

bob bundy
2013-10-03 00:07:41

So,WHO IS RIGHT!!!

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

Bob

bobbym
2013-10-02 22:29:20

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

auyeungyat
2013-10-02 22:25:29

So,WHO IS RIGHT!!!

bobbym
2013-10-02 21:54:07

Hi;

I am getting:

P(at most one 6) = 25 / 27

P(at least 2 sixes) = 2 / 27

done by the binomial distribution.

auyeungyat
2013-10-02 21:46:45

It is wrong because
1.P=[1/6(six)x5/6(not six)x5/6(not six)=25/216]
Can you think that 191/216 is right?
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]

bob bundy
2013-10-02 09:00:04

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

bob bundy
2013-10-02 08:53:10

hi 14_karat

Stay on-line while I put together an explanation.

Bob

14_karat
2013-10-02 08:45:58

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!