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Please help with Probabilty problems!!!

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Please help with Probabilty problems!!! [#permalink] New post 23 May 2003, 11:21
Hi,
I was unable to solve these 3 problems from your probability lesson, please can you help me:

3. How many different signals can be transmitted by hoisting 3 red, 4 yellow and 2 blue flags on a pole, assuming that in transmitting a signal all nine flags are to be used? (Ans. 1260)

I would say that it should be done as a problem where you have to find combinations of different words that can be formed from 9 letters where 3 letters A, 4 letters B and 2 letters C. In this case it should be 9!/(3!*4!*2!), but I do not fully understand a logic behind this?


2. There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group? (Ans. 80)

There are C(10,3)=120 combinations, minus combinations when there is a husband and a wife in a group of three: 5(5 different couples)*8(8 choices for one more place in a group)=40. Total 120-40=80. Is it a right approach?

7. The probability that it will rain in NYC on any given day in July is 30%. What is the probability that it will rain on exactly 3 days from July 5 to July 10 ?

There are C(5,3)=10 desired combinations of rain in 3 days and it should be 32 possible combinations if probability of rain is 1/2, but we have probability of rain as 30%, i do not know what is the next step?

Thanks, Boris
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Re: Please help with Probabilty problems!!! [#permalink] New post 23 May 2003, 12:16
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My comments are in BLUE...


Hi,
I was unable to solve these 3 problems from your probability lesson, please can you help me:

3. How many different signals can be transmitted by hoisting 3 red, 4 yellow and 2 blue flags on a pole, assuming that in transmitting a signal all nine flags are to be used? (Ans. 1260)

I would say that it should be done as a problem where you have to find combinations of different words that can be formed from 9 letters where 3 letters A, 4 letters B and 2 letters C. In this case it should be 9!/(3!*4!*2!), but I do not fully understand a logic behind this?

A good way to approach Permutations, etc is to build a model - a simple one. Take 4 flags total, 3 red and 1 blue and try to work with them. Basically, what you are doing by dividing 9! by 3!4!2! is you omitting the "internal" combinations - that is when the change happened within the 3 red flags, which is not noticeable. 3! is in how many ways you can arrange 3 different flags, but since our flags the same, all of these combinations have to be extracted (not subtracted since there is a number of other players - therefore division).



2. There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group? (Ans. 80)

Here is a discussion about this topic; it was a pretty lengthy one; I hope it is clear. Post more if not.
http://www.gmatclub.com/phpbb/viewtopic.php?t=213&highlight=husband


There are C(10,3)=120 combinations, minus combinations when there is a husband and a wife in a group of three: 5(5 different couples)*8(8 choices for one more place in a group)=40. Total 120-40=80. Is it a right approach?

7. The probability that it will rain in NYC on any given day in July is 30%. What is the probability that it will rain on exactly 3 days from July 5 to July 10 ?

There are C(5,3)=10 desired combinations of rain in 3 days and it should be 32 possible combinations if probability of rain is 1/2, but we have probability of rain as 30%, i do not know what is the next step?

Here are explanations of this problem:

http://www.gmatclub.com/phpbb/viewtopic.php?t=100&highlight=rain

http://www.gmatclub.com/phpbb/viewtopic.php?t=705&start=0&postdays=0&postorder=asc&highlight=rain

http://www.gmatclub.com/phpbb/viewtopic.php?t=182&highlight=rain

Thanks, Boris
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 [#permalink] New post 23 May 2003, 14:28
Thanks a lot for your help. The only problem that still is not 100% clear form me is a question about rain. This problem is resolved in 2 different ways:

1. P(3 rainy days and 3 nonrainy) = 0.3*0.3*0.3*0.7*0.7*0.7*20=0.18522

2. Probability of raining on three days in a five day period=5c3(0.3)(0.3)(0.3)(0.2)(0.2)

Why in second example we multiply by 0.2 and not by 0.7?

boris
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 [#permalink] New post 23 May 2003, 15:13
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bobzilb wrote:
Thanks a lot for your help. The only problem that still is not 100% clear form me is a question about rain. This problem is resolved in 2 different ways:

1. P(3 rainy days and 3 nonrainy) = 0.3*0.3*0.3*0.7*0.7*0.7*20=0.18522

2. Probability of raining on three days in a five day period=5c3(0.3)(0.3)(0.3)(0.2)(0.2)

Why in second example we multiply by 0.2 and not by 0.7?

boris



Here is the correct answer: 6! / (6 - 3)!3! x 0.3 x 0.3 x 0.3 x 0.7 x 0.7 x 0.7 .

It is 6 days, so C(6,3)=20, which is the same as your answer #1. I am not sure about 0.2*0.2 I don't know where that came from and the answer does not match... probably somebody made a mistake :)


---
  [#permalink] 23 May 2003, 15:13
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