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This problem is from GMATClub: Five coins are tossed one

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This problem is from GMATClub: Five coins are tossed one [#permalink] New post 08 Jan 2004, 03:17
This problem is from GMATClub:

Five coins are tossed one after the other. What is the probability that the first three are heads?

A. 1/16
B. 5/16
C. 1/2
D. 2/3
E. 11/16

Can somebody explain to me why the answer is B? :wall
Thanks in advance.
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 [#permalink] New post 08 Jan 2004, 03:48
The prob of 3 heads out of five is 5C3x(1/2)^5= 10/32=5/16. , you can also use binominal formula (1/2+1/2)^5. Since the events are independent the probability of first tree being heads is the same as probability of any 3 coins being heads out of 5.
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 [#permalink] New post 08 Jan 2004, 06:46
I don't understand your logic.

This is the way I thought of it:

The first three coins have to land on heads and the last two can land on either heads or tails. Hence, here is the list of possibilities:

HHHHH
HHHHT
HHHTH
HHHTT

That's four ways out of a total of 2^5 possible ways to flip the coins. So the answer I got was 4/32=1/8.

Even though 1/8 is not an answer choice, I still think it's the correct answer. Any input?
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 [#permalink] New post 08 Jan 2004, 07:45
I think you are right that P=1/8.
The question is ambiguous and I believe your interpretation
of the question is correct.
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 [#permalink] New post 31 Mar 2005, 20:15
Can someone explain the answer to this one again? I am not sure why the answer is not 1/8...
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 [#permalink] New post 31 Mar 2005, 21:49
Total number of possible combinations:
0H, 5T - 1 way
1H, 4T - 5 ways
2H, 3T - 10 ways
3H, 2T - 10 ways
4H, 1T - 5 ways
5H, 0T - 1 way

Total number of combinations = 32

Winning combinations: HHHTT, HHHHT, HHHTH, HHHHH = 4 combinations

But we only have 1 winning combination, therefore probability is 4/32 = 1/8
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Re: Probability - Coin toss problem [#permalink] New post 01 Apr 2005, 09:15
If the question is this:

Quote:
Five coins are tossed one after the other. What is the probability that the exactly three are heads?


Then the answer is 5/16.
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 [#permalink] New post 01 Apr 2005, 10:44
we could definitely use the binomial formula

5c3*(1/2)^3*(1/2)^2 -> 10/32
nCr*(p)^r*(1-p)^(n-r)
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Reply [#permalink] New post 14 Jul 2005, 00:39
I think the answer is 5/16

Sample space= 2^5= 32
Three heads out of 5= 5C3

5C3/ 32= 10/ 32= 5/16

Thanks
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 [#permalink] New post 15 Jul 2005, 16:59
http://www.gmatclub.com/content/courses/quantitative/probability.php
explains binominal formula.

But I think Binomial Distribution only checks a certain results happens K times out N trials. In other words, 3 tosses have heads up out of 5 tosses,
not necessarily "the first three tosses have heads up"

Is the question the same as throwing a coin 3 times and expect
each time it has heads up ?

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 [#permalink] New post 16 Jul 2005, 03:03
This questions doesn't explain what it means to flip 5 coins, but only calculate the probability of 3 of them. Are GMAT questions ambiguous like this? Or will they be better defined?
  [#permalink] 16 Jul 2005, 03:03
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This problem is from GMATClub: Five coins are tossed one

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