ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: schnappy on November 07, 2010, 12:58:05 pm
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The sample space when a fair twelve-sided die is rolled is {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}. Each outcome
is equally likely.
For which one of the following pairs of events are the events independent?
A. {1, 3, 5, 7, 9, 11} and {1, 4, 7, 10}
B. {1, 3, 5, 7, 9, 11} and {2, 4, 6, 8, 10, 12}
C. {4, 8, 12} and {6, 12}
D. {6, 12} and {1, 12}
E. {2, 4, 6, 8, 10, 12} and {1, 2, 3}
I understand that Pr(A)*Pr(B) = Pr(A [upside U] B)
But how do they know straight off the bat that Pr(A [upside U] B) = 1/6?
Thanks
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For A, {1,7} are the only 2 that is similar for both events, therefore, 2/12 gives 1/6.
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yeh jinxi is right.. th answer would be A..
if u make x = {1, 3, 5, 7, 9, 11} and y = {1, 4, 7, 10}
total numbers = 12
pr(x) = 6/12
pr(y) = 4/12
pr( x intersect y ) = 2/12 because they have 2 numbers common {1,7}
so 2/10 = 6/12 * 4/12
simplify u get 1/6 = 1/6 so it is independant
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yeh jinxi is right.. th answer would be A..
if u make x = {1, 3, 5, 7, 9, 11} and y = {1, 4, 7, 10}
total numbers = 10
pr(x) = 6/10
pr(y) = 4/10
pr( x intersect y ) = 2/10 because they have 2 numbers common {1,7}
so 2/10 = 6/10 * 4/10
simplify u get 1/5 = 1/5 so it is independant
6/10 * 4/10 does not give 1/5 ^^
How i do it is since,
x = {1, 3, 5, 7, 9, 11} and y = {1, 4, 7, 10}, and Total Sample space = 12.
Event A has 6/12, Event B has 4/12,
6/12*4/12 = 1/6
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You just have to work through the options. Write the probablility of each set just above it and the probability of the intersection to the right. Then just go through A, B, C etc. until you find the answer.
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yeh jinxi is right.. th answer would be A..
if u make x = {1, 3, 5, 7, 9, 11} and y = {1, 4, 7, 10}
total numbers = 10
pr(x) = 6/10
pr(y) = 4/10
pr( x intersect y ) = 2/10 because they have 2 numbers common {1,7}
so 2/10 = 6/10 * 4/10
simplify u get 1/5 = 1/5 so it is independant
6/10 * 4/10 does not give 1/5 ^^
LOL!!! my bad .. soz :S
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How i do it is since,
x = {1, 3, 5, 7, 9, 11} and y = {1, 4, 7, 10}, and Total Sample space = 12.
Event A has 6/12, Event B has 4/12,
6/12*4/12 = 1/6
is this correct because 1/6 is equal to the probability of a intersection ( 1/6 due to there being 2 shared numbers).
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How i do it is since,
x = {1, 3, 5, 7, 9, 11} and y = {1, 4, 7, 10}, and Total Sample space = 12.
Event A has 6/12, Event B has 4/12,
6/12*4/12 = 1/6
is this correct because 1/6 is equal to the probability of a intersection ( 1/6 due to there being 2 shared numbers).
Yeah.
Pr(A)xPr(B) must equal the probability of intersection by counting.
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cheers