3 questions in the tradition of Stolyar... Three fair dice

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3 questions in the tradition of Stolyar... Three fair dice [#permalink]

09 Dec 2003, 08:07

3 questions in the tradition of Stolyar...

Three fair dice are rolled, what is the probability that

(1) you get 3 sixes?
(2) you DO NOT get 3 sixes?
(3) you get ATLEAST one six?

SVP

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09 Dec 2003, 08:15

in the tradition... looks great!!! nice to see it!!!

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09 Dec 2003, 08:49

Quote:

(1) you get 3 sixes?

(1/6)^3

Quote:

(2) you DO NOT get 3 sixes?

1-((1/6)^3)

Quote:

(3) you get ATLEAST one six?

1-((5/6)^3)

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09 Dec 2003, 09:44

Let's increase the difficulty a bit, what is the probability that

you will NOT get EXACTLY 2 sixes?

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09 Dec 2003, 11:52

Quote:

Let's increase the difficulty a bit, what is the probability that

you will NOT get EXACTLY 2 sixes?

x66 (5 possibilities)
6x6 (5 possibilities)
66x (5 possibilities)

15 ways out of 6^3 to get exactly 2 sixes.

((6^3)-15)/6^3

Last edited by stoolfi on 09 Dec 2003, 17:49, edited 1 time in total.

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09 Dec 2003, 15:48

Is it 25 or 15? (3 x 5)

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09 Dec 2003, 17:48

My mistake. I edited my errant message.

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Re: Dice probability [#permalink]

13 Dec 2003, 02:02

bluefox420 wrote:

3 questions in the tradition of Stolyar...

Three fair dice are rolled, what is the probability that

(1) you get 3 sixes?
(2) you DO NOT get 3 sixes?
(3) you get ATLEAST one six?

Quote:

(1) you get 3 sixes?

only one way out of 216 ways => 1/216

Quote:

(2) you DO NOT get 3 sixes?

1- 1/216 =215/216

Quote:

(3) you get ATLEAST one six?

[/quote]

1- (5/6)^3 = 91/216

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13 Dec 2003, 02:05

bluefox420 wrote:

Let's increase the difficulty a bit, what is the probability that you will NOT get EXACTLY 2 sixes?

P (Exactly two sixes) = 3 * 1/6^2 * 5/6 = 5/72

Reqd prob = 1 - 5/72 = 67/72

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13 Dec 2003, 02:06

stoolfi wrote:

Quote:

Let's increase the difficulty a bit, what is the probability that

you will NOT get EXACTLY 2 sixes?

x66 (5 possibilities)
6x6 (5 possibilities)
66x (5 possibilities)

15 ways out of 6^3 to get exactly 2 sixes.

((6^3)-15)/6^3

SVP

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05 Jan 2004, 17:16

Hi stolyar

for the improved question you need to add getting 3 sixes to getting 1 six.

x66
6x6
66x
666

1/6^3 + 3 * 1/6 * 5/6 * 5/6

SVP

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10 Jan 2004, 15:33

My bad. stoolfi is right on the improved question. Q is asking not getting exactly 2 sixes.

[#permalink] 10 Jan 2004, 15:33

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3 questions in the tradition of Stolyar... Three fair dice

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3 questions in the tradition of Stolyar... Three fair dice

3 questions in the tradition of Stolyar... Three fair dice

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