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 November 14th, 2011, 11:19 AM #1 Newbie   Joined: Oct 2011 Posts: 12 Thanks: 0 probability dice If you roll two fair dice, one red and one green. Whatis the probability that you get a sum of 7 given that one die is an even number? Thank you.
 November 14th, 2011, 11:41 AM #2 Senior Member   Joined: Oct 2011 From: Belgium Posts: 522 Thanks: 0 Re: probability dice 6 possibilities for 7 with 1 die even 1 + 6 6 + 1 4 + 3 3 + 4 5 + 2 2 + 4 possibilities with at least 1 die even = 3*3+3*3+3*3 = 27 answer: 6/27
 November 14th, 2011, 11:54 AM #3 Newbie   Joined: Oct 2011 Posts: 12 Thanks: 0 Re: probability dice sorry, I dont get it why possibilities with at least 1 die even = 3*3+3*3+3*3 = 27?
November 14th, 2011, 12:07 PM   #4
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Re: probability dice

Quote:
 Originally Posted by mathproblems sorry, I dont get it why possibilities with at least 1 die even = 3*3+3*3+3*3 = 27?
first die even, second die uneven: 3*3
second die even, first die uneven: 3*3
first die even, second die even: 3*3

 November 14th, 2011, 12:12 PM #5 Newbie   Joined: Oct 2011 Posts: 12 Thanks: 0 Re: probability dice sorry, can did you get 3?
November 14th, 2011, 12:30 PM   #6
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Re: probability dice

Quote:
 Originally Posted by mathproblems sorry, can did you get 3?
because there are 3 even and 3 uneven values on a die

November 15th, 2011, 06:58 AM   #7
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Re: probability dice

Hello, mathproblems!

Quote:
 You roll two fair dice, one red and one green. What is the probability that you get a sum of 7, given that one die is an even number?

$\text{There are: }\,6\,\times\,6 \:=\:36\text{ possible outcomes.}$

$\text{There are: }\,3\,\times\,3\:=\:9\text{ outcomes with both numbers odd.}$

$\text{Hence, there are: }\:36\,-\,9\:=\:27\text{ outcomes with at least one number even.}$

$\text{Among them, there are }six\text{ with a sum of 7: }\1,\,6),\2,\,5),\3,\,4),\4,\,3),\5,\,2) ,\6,\,1)" />

$\text{Therefore: }\:P(\text{sum of 7 }|\text{ at least one even}) \:=\:\frac{6}{27} \:=\:\frac{2}{9}$

 November 15th, 2011, 07:38 AM #8 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: probability dice If you are given that one die is even, it means that precisely one die is even right? Meaning the other is odd. 18 possibilities to have precisely 1 even. So a 1/3 chance that it'll give a 7.
 November 15th, 2011, 07:40 AM #9 Newbie   Joined: Oct 2011 Posts: 12 Thanks: 0 Re: probability dice thank you. sounds great
 November 16th, 2011, 10:08 PM #10 Newbie   Joined: Oct 2011 From: Jaipur Posts: 17 Thanks: 0 Re: probability dice Specific Multiplication Rule Only valid for independent events P(A and B) = P(A) * P(B) Example 3: P(A) = 0.20, P(B) = 0.70, A and B are independent. B B' Marginal A 0.14 0.06 0.20 A' 0.56 0.24 0.80 Marginal 0.70 0.30 1.00 The 0.14 is because the probability of A and B is the probability of A times the probability of B or 0.20 * 0.70 = 0.14. Dependent Events If the occurrence of one event does affect the probability of the other occurring, then the events are dependent. Conditional Probability The probability of event B occurring that event A has already occurred is read "the probability of B given A" and is written: P(B|A) General Multiplication Rule Always works. P(A and B) = P(A) * P(B|A)

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