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# this will probably be fairly easy for some of you, but i'm

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this will probably be fairly easy for some of you, but i'm [#permalink]

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26 Jul 2004, 12:03
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

this will probably be fairly easy for some of you, but i'm stuck on it:

a fair coin with sides marked heads and tails is to be tossed eight times. what is the probability that the coin will land tails side up more than five times?

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26 Jul 2004, 12:18
We would have to use the bernoullis theorem....

P=nPk+nP(k+1)+.....nPn

where nPk=nCk*(p^k)*((1-p)^(n-k))....sounds simple and fair enuff

Lets calculate....
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26 Jul 2004, 12:27
Karthik wrote:
We would have to use the bernoullis theorem....

P=nPk+nP(k+1)+.....nPn

where nPk=nCk*(p^k)*((1-p)^(n-k))....sounds simple and fair enuff

Lets calculate....

ah, i c. thanks, this helps a lot.

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26 Jul 2004, 13:06
ian,

is there any easy way of handling this....calculating the probability in this case is cumbersome...is this a GMAT Type anyway??
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26 Jul 2004, 14:24
IMO, formulas are good, but not good enough. do try to understand why the formula works the way it works. ETS makes a living out of making simple problems look difficult. If just one thing is out of place, it wouldnt be difficult to panic.

Search for explanations by anupag/akamai brah/stolyar for this.

Thanks
Praetorian

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26 Jul 2004, 19:30
so is the final ans (1/2)^8 ( 8C5 + 8C6 + 8C7 + 8C8)
= 93/256

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Location: Russian Federation
Concentration: Entrepreneurship, International Business
GMAT 3: 740 Q40 V50
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WE: Supply Chain Management (Energy and Utilities)

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26 Jul 2004, 22:49
Way to approach the problem is as follows:

1. Total no. of outcomes is 2^8. You are flipping 8 times and probability of tail showing up everytime you flip the coin is 1/2. When you do it eight times it is 1/2^8. Or the total outcomes (denominator) is 2^8.

2. Now, the question is what is the probability of tails coming up atleast 5 times - this means that tails can appear 5 times, 6 times, 7 times or all the 8 times. Think combinations and you'll get 8C5 + 8C6 + 8C7 + 8C8 ways of getting atleast 5 tails and this adds up to 93.

3. The answer would be 93/2^8 = 93/256.

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Director
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26 Jul 2004, 23:41
Sorry did not log in last time. the guest was me.

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27 Jul 2004, 06:41
Karthik wrote:
ian,

is there any easy way of handling this....calculating the probability in this case is cumbersome...is this a GMAT Type anyway??

I agree with smandalika and venksune on this one. However, the question asks for the "probability that the coin will land tails side up more than five times". I don't think that includes 5 times, so I only did it with 6, 7, and 8 times. My answer: 37/256.

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27 Jul 2004, 20:44
good catch Ian. I somehow missed that one

regards

smandalika

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27 Jul 2004, 23:21
Ian is right. Should be 37/256.

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28 Jul 2004, 14:02
ian7777 wrote:
Karthik wrote:
ian,

is there any easy way of handling this....calculating the probability in this case is cumbersome...is this a GMAT Type anyway??

I agree with smandalika and venksune on this one. However, the question asks for the "probability that the coin will land tails side up more than five times". I don't think that includes 5 times, so I only did it with 6, 7, and 8 times. My answer: 37/256.

indeed, 37/256 is the OA.

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28 Jul 2004, 14:02
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# this will probably be fairly easy for some of you, but i'm

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