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A fair coin with sides marked heads and tails is to be

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A fair coin with sides marked heads and tails is to be [#permalink] New post 02 Mar 2005, 10:16
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?Please explain this for me.
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 [#permalink] New post 02 Mar 2005, 10:22
8 times => 8c8 * 256 = 1/256
7 times => 8c7 * 256 = 8/256
6 times => 8c6 * 256 = 28 / 256

=> 37/256
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 [#permalink] New post 02 Mar 2005, 10:50
Probability of getting tails eight times = 1/256
Probability of getting tails seven times = 1/128
Probability of getting tails six times = 1/64
So Probability of getting tails more than five times = 1/64 + 1/128 + 1/256 = 7/256
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 [#permalink] New post 02 Mar 2005, 12:59
I think it should be:

Prob of 8 tails= 0.2^8 =

Prob of 7 tails= C(8,7)*0.2^7

Prob of 6 tails= C(8,6)*0.2^6

is my reasoning right??
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 [#permalink] New post 02 Mar 2005, 14:00
christoph wrote:
8 times => 8c8 * 256 = 1/256
7 times => 8c7 * 256 = 8/256
6 times => 8c6 * 256 = 28 / 256

=> 37/256

I totally agree with christoph
You have (1/2)^8 ways of getting a different result (H or T). Then with each of the outcomes(>5... which is 6,7 and 8), you have 8C6, 8C7 and 8C8 ways of sorting the H possibilities.
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 [#permalink] New post 02 Mar 2005, 14:04
kevinw wrote:
I think it should be:

Prob of 8 tails= 0.2^8 =

Prob of 7 tails= C(8,7)*0.2^7

Prob of 6 tails= C(8,6)*0.2^6

is my reasoning right??

I'm not sure where you got the 0.2 from. You should use the following formula which christoph used: sum of[nCr (p)^r (q)^(n-r)]
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 [#permalink] New post 02 Mar 2005, 14:07
I initially thought it was 7/8 since I multiplied the three situations of 6,7 and 8..

however, they are all three independent so should be added.

[Dependent cases are multiplied while independent are added. ] Right?

Answer 37/256.
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 [#permalink] New post 02 Mar 2005, 14:10
bozo, you are right on!
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 [#permalink] New post 02 Mar 2005, 14:18
kevinw wrote:
I think it should be:

Prob of 8 tails= 0.2^8 =

Prob of 7 tails= C(8,7)*0.2^7

Prob of 6 tails= C(8,6)*0.2^6

is my reasoning right??


No. First, 0.2 should be 0.5 (it might be a typo).

However,
Prob of 7 tails= C(8,7)*0.5^7*0.5
Prob of 6 tails= C(8,6)*0.5^6*0.5^2

Remember the formula have p^r*(1-p)^(n-r).
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 [#permalink] New post 02 Mar 2005, 15:19
:oops: :oops: :oops:

I suck!!!.......

Ms. Hong, you are right....(as usual)...
and I am wrong, as usual ...

thank you for the lesson!...
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 [#permalink] New post 02 Mar 2005, 15:34
kevinw wrote:
and I am wrong, as usual ...


Don't be so harsh on yourself. You just wasn't as used to the formula as I am. From now on you'll be always right on this type of questions, won't you? ;)
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 [#permalink] New post 02 Mar 2005, 17:32
Heres my 2 cents on these type of problems

Fair Coin -------> probablity of heads or tails = 1/2

You need to pick more than 5 instances out of 8 where it is heads/tails(it doesnt matter as it is a fair coin)

So it should be 6,7 or 8 times the desired outcome may arrive

Take 6 times out of 8 then the chaces of heads/tails
= 8C6 (1/2)^6*(1/2)^2

that is nCr*P(E)^a*P(E!)^b

Where n = no of total outcomes
r = no of desired outcomes
P(E) is probablity of desired outcomes
a -------> no of desired outcomes
P(E!) is probablity of UNdesired outcomes
b -------> no of UNdesired outcomes

n=a+b

Since we have r=6,7,8 add them all after putting in the above formula


Thanks
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 [#permalink] New post 02 Mar 2005, 17:34
I think Tiger is a very lucky boy, to have such a comprehensive mom...

I don't want to say my mom wasn't good, but I never got this kind of support...... :cry: :cry: :cry:
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Re: Probability [#permalink] New post 02 Mar 2005, 18:36
Grant wrote:
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?Please explain this for me.



Prob of 8 tails= C(8,8)*0.5^8
Prob of 7 tails= C(8,7)*0.5^7*0.5
Prob of 6 tails= C(8,6)*0.5^6*0.5^2


= 0.5^8(C(8,8)*C(8,7)*C(8,6)
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 [#permalink] New post 03 Mar 2005, 02:18
obviously p=0.5 not 0.2 :oops:
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 [#permalink] New post 03 Mar 2005, 11:20
kevinw wrote:
Tiger


:eyes
You are so thorough. ;);)
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 [#permalink] New post 03 Mar 2005, 13:00
I almost got this right until i multiplied instead of adding.
Is there an exclusive rule as to when to Add or Multiply?

I know when prob or x AND y, we multiply
I know also that for prob x OR y, we Add


but how else do we know when to Add or multiply? Help pls. Thanks :cry:
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 [#permalink] New post 03 Mar 2005, 13:11
HongHu wrote:
kevinw wrote:
I think it should be:

Prob of 8 tails= 0.2^8 =

Prob of 7 tails= C(8,7)*0.2^7

Prob of 6 tails= C(8,6)*0.2^6

is my reasoning right??


No. First, 0.2 should be 0.5 (it might be a typo).

However,
Prob of 7 tails= C(8,7)*0.5^7*0.5
Prob of 6 tails= C(8,6)*0.5^6*0.5^2

Remember the formula have p^r*(1-p)^(n-r).


HongHu, please explain this formula (P^r) * (1-P)^(n-r) and how and when to apply it to permutation, comb and prob. problems. Thanks
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will this type of question be on GMAT? [#permalink] New post 05 Mar 2005, 08:23
Since you're discussing this, I'm assuming it will. But it does seem a bit hard for the GMAT? or maybe I'm completely wrong.

Thanks,

Mike
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 [#permalink] New post 05 Mar 2005, 09:43
This is one of the easiest you can have in GMAT if it comes
  [#permalink] 05 Mar 2005, 09:43
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