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# Probability of repeated trials

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Senior Manager
Joined: 31 Jul 2006
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Probability of repeated trials [#permalink]  08 Aug 2006, 16:51
guys I am looking for a formula to solve this problem:

Coin tossed 4 times. What is probability that atleast 3 out of 4 tosses will come up heads? (answer is 5/16)

kaplan did it by writing out all combinations of H&Ts. I know that's going to be a PITA on exam day for me, so I'd rather have a neat formula handy.

Why doesn't this formula work for me?

.....................N!
BDF = ------------------ * p^M * (1 â€” p)^N â€” M
.............(N â€” M)! * M!

BDF = the binomial distribution probability;
p = the individual probability of the phenomenon (p = 0.5 to get heads in coin tossing);
M = the exact number of successes
N = the number of trials

Thx!

Last edited by Nsentra on 08 Aug 2006, 16:55, edited 1 time in total.
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Manager
Joined: 08 May 2004
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I have the same problem! How in the heck are you going to right it all out on exam day without wasting 2-3 minutes on a problem?
I have never seen it asked like that on GMAT prep..
so I wouldln't worry too too much..
Senior Manager
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jamesrwright3 wrote:
I have the same problem! How in the heck are you going to right it all out on exam day without wasting 2-3 minutes on a problem?
I have never seen it asked like that on GMAT prep..
so I wouldln't worry too too much..

I appreciate Kaplan for putting some stuff for me in baby terms, but sometimes by not explaining everything behind the answer they are preventing me from seeing the real math behind the solution
Manager
Joined: 26 Jun 2006
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Why am not getting 5/16? In fact, I am getting 1-5/16=11/16... Confused

Pron(0 or 1 heads) = ?

Total number of possible combinationsL 2^4=16

0 heads: 1 combination (when all tails)
1 head, 3 tails: 4 combinations

Prob(0 or 1) = (1+4)/16 = 5/16

Prob(3 or more)=1-5/16=11/16

???
Senior Manager
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v1rok wrote:
Why am not getting 5/16? In fact, I am getting 1-5/16=11/16... Confused

Pron(0 or 1 heads) = ?

Total number of possible combinationsL 2^4=16

0 heads: 1 combination (when all tails)
1 head, 3 tails: 4 combinations

Prob(0 or 1) = (1+4)/16 = 5/16

Prob(3 or more)=1-5/16=11/16

???

They are saying there are 5 possible outcomes with alteast 3 heads in 4 tosses

hhht
hhth
hthh
thhh
hhhh

the word 'atleast' is prob. why my probability formula is not working out

so 5/16

Last edited by Nsentra on 08 Aug 2006, 17:20, edited 1 time in total.
Manager
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Nevermind, I found where I got derailed:

P(3 or more)= 1 - P(0,1,or2)

He-he, looks like I need some rest...
Senior Manager
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v1rok wrote:
Nevermind, I found where I got derailed:

P(3 or more)= 1 - P(0,1,or2)

He-he, looks like I need some rest...

Is this the only way to solve this problem? Write out all possible outcomes? there is no algebraic formula?
Manager
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Exactly, nissan! I was working too quickly (trying to fit in those 2 minutes!) and accidently skipped one case...

Anyway, you don't have to write down all possible combinations of heads and tails.

You just need to remember that 2^4 is total number of combinations (=16)
and then count only those cases where we have eaither 4 heads (1 such case) or 3 heads (4 cases). So,

Prob=5/16
Senior Manager
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v1rok wrote:
Exactly, nissan! I was working too quickly (trying to fit in those 2 minutes!) and accidently skipped one case...

I know for a fact that is what is going to happen to me on a test. Thanks for giving another thought on how to do this though.
Manager
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Nsentra wrote:
v1rok wrote:
Exactly, nissan! I was working too quickly (trying to fit in those 2 minutes!) and accidently skipped one case...

I know for a fact that is what is going to happen to me on a test. Thanks for giving another thought on how to do this though.

Cheer up! That's exactly why you are here, isn't it. A little bit of practicing and you will get confident enough so that none of the tricks GMAT trows at you would make you blink...
CEO
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Prob of getting exactly three heads = 4C3 * (1/2)^3 * (1/2)^1 = 4/16

Prob of getting exactly 4 heads = 4C4 * (1/2)^4 = 1/16

Total prob = 1/16 + 4/16 = 5/16
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

Last edited by ps_dahiya on 11 Aug 2006, 10:18, edited 1 time in total.
Senior Manager
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ps_dahiya wrote:
Prob of getting exactly three heads = 4C3 * (1/2)^3 * (1/2)^1 = 4/16

Prob of getting exactly 4 heads = 4C4 * (1/2)^4 = 1/16

Total prob = 1/16 + 4/16 = 5/16

Perfect! Thanks! (great quote too)
Manager
Joined: 03 Jul 2006
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ps_dahiya,

could you explain the formula that you used. seems simple
CEO
Joined: 20 Nov 2005
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rkatl wrote:
ps_dahiya,

could you explain the formula that you used. seems simple

Its is explained here by the math goddess HongHu:

http://www.gmatclub.com/phpbb/viewtopic.php?t=14706
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

Director
Joined: 17 Jul 2006
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PS_dahiya,
There is a typo in your reply. Please don't mistake me. I know you are so good. Just to help others to understand ....

Probability of winning head or tail in tossing coin is 1/2

Prob of getting exactly three heads = 4C3 * (1/2)^3 * (1/3)^1 = 4/16

Prob of getting exactly 4 heads = 4C4 * (1/2)^4 = 1/16

Total prob = 1/16 + 4/16 = 5/16.

Manager
Joined: 12 May 2006
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How I approached the problem:

16 possible outcomes altogether (4*4)

Getting at least 3:
Getting all 4-
4C4 = 1

Getting 3 out of 4:
4C3 = 4

4+1 = 5

5/16 [/u]
CEO
Joined: 20 Nov 2005
Posts: 2913
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
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Kudos [?]: 121 [0], given: 0

PS_dahiya,
There is a typo in your reply. Please don't mistake me. I know you are so good. Just to help others to understand ....

Probability of winning head or tail in tossing coin is 1/2

Prob of getting exactly three heads = 4C3 * (1/2)^3 * (1/3)^1 = 4/16

Prob of getting exactly 4 heads = 4C4 * (1/2)^4 = 1/16

Total prob = 1/16 + 4/16 = 5/16.

Thanks buddy. I have corrected the error.
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

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