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To be attested to a black belt, a brown-belt karatist has to

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To be attested to a black belt, a brown-belt karatist has to [#permalink] New post 19 Dec 2003, 02:19
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To be attested to a black belt, a brown-belt karatist has to go through six sparrings. He has to win all three figths with brown belts and then at least one out of three figths with black belts. His probability to win a brown belt is 0.5; his probability to win a black belt is 0.3. What is the probability that the guy gets a black belt?
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 [#permalink] New post 19 Dec 2003, 04:27
0.125x3x0.3+0.125x0.09x3+0.125x0.027=1197/8000
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 [#permalink] New post 19 Dec 2003, 04:33
BG wrote:
0.125x3x0.3+0.125x0.09x3+0.125x0.027=1197/8000


0.125[(0.3*0.3*0.3)+(0.3*0.3*0.7)+(0.3*0.7*0.7)]=0.125(0.027+0.063+0.147)=0.125*0.237=~0.03
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 [#permalink] New post 19 Dec 2003, 04:47
stolyar wrote:
BG wrote:
0.125x3x0.3+0.125x0.09x3+0.125x0.027=1197/8000


0.125[(0.3*0.3*0.3)+(0.3*0.3*0.7)+(0.3*0.7*0.7)]=0.125(0.027+0.063+0.147)=0.125*0.237=~0.03


It is not exactly so. You considered only: winning three black belts, then first two black belts and first blackbelt. What about three other possible cases: probabilities of winning 1)only second blackbelt 2) only third 3)second and third.

Actually, instead of considering so many cases, one needs to condsider only the probability of not loosing to all three balckbelts:, which is 1-0.7*0.7*0.7


Final answer will be: 0.125*[1-0.7*0.7*0.7] = whatever it is
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 [#permalink] New post 19 Dec 2003, 09:12
agree with your approach

I should have written:
0,125*[[1С3*0.3*0.7*0.7]+[2C3*0.3*0.3*0.7]+[3C3*0.3*0.3*0.3]]

I have influenza and therefore pretty stupid...
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 [#permalink] New post 05 Jan 2004, 14:46
is the following correct

Pbr - probability that he wins brown belt match
Pbw - probability that he wins black belt match
Pbl - probability that he loses black belt match

So probability that he gets black belt =

Pbr Pbr Pbr Pbw Pbl Pbl +
Pbr Pbr Pbr Pbl Pbw Pbl +
Pbr Pbr Pbr Pbl Pbl Pbw +
Pbr Pbr Pbr Pbw Pbw Pbw


0.5 ^ 3 ( 3 * 0.3 * 0.7 ^ 2 + 0.3 ^ 3 )
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 [#permalink] New post 05 Jan 2004, 18:49
anvar wrote:
Actually, instead of considering so many cases, one needs to condsider only the probability of not loosing to all three balckbelts:, which is 1-0.7*0.7*0.7


Final answer will be: 0.125*[1-0.7*0.7*0.7]


The best approach to this kind of probability problems. I have seen at least three times in last two days similar problems. eg What is the probability that a head of coin will be result at least once from 6 throws...or at least 2 times.....It takes less time to caculate probability for 0 and for 1 head than to sum up probability for results with 6 , 5 ,4 ,3 and 2 heads . It is very useful to use "reverse" logic.
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 [#permalink] New post 05 Jan 2004, 19:16
Sorry. The problem said atleast one win with black belt person. So reverse method is the shortest.
  [#permalink] 05 Jan 2004, 19:16
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