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junior college 1 | H2 Maths
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Help to find solution please
Date Posted: 6 years ago
Views: 420
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Probability of one combination (XXXYor Z)=200/430*199/429*198/428*230/427
There are 4 possible such combinations.
Hence probability of three winners from X will be = 4*200/430*199/429*198/428*230/427
There are 4 possible such combinations.
Hence probability of three winners from X will be = 4*200/430*199/429*198/428*230/427
Date Posted:
6 years ago
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B) probability all winners coming from Y = (50*49*48*47)/(430*429*428*427)
Date Posted:
6 years ago
Sorry it should be (80*79*78*77)/(430*429*428*427)
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This is conditional probability.
To find this obtain Probability of three winners from X and one winner from Z divided that three winners are from X
P(3X and Z) = 4*200/430*199/429*198/428*150/427
P(Three winners from X)= answer in part a
Hence P(3X and Z/3X)= 150/230=15/23
To find this obtain Probability of three winners from X and one winner from Z divided that three winners are from X
P(3X and Z) = 4*200/430*199/429*198/428*150/427
P(Three winners from X)= answer in part a
Hence P(3X and Z/3X)= 150/230=15/23
Date Posted:
6 years ago
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