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Five coins are tossed one after the other. What is the

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Five coins are tossed one after the other. What is the [#permalink] New post 24 Mar 2003, 21:11
Five coins are tossed one after the other. What is the probability that the first three are heads.

I hope I have the framed the question properly. :peek

don't :beat me

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 [#permalink] New post 24 Mar 2003, 22:47
P(HHHTT)=5C3/32=10/32=5/16

Total outcomes: 2^5=32
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nope [#permalink] New post 18 May 2003, 06:17
Stolyar, once again i disagree :roll:

p(First is heads) = 1/2
p(Second is heads) = 1/2
p(Third is heads) = 1/2

Events are independent, so p(ABC) = p(A)p(B)p(C)

p = 1/2 ^ 3 = 1/8!

In your case you forget that HHHHT and HHHTH also qualify as favorable. It doesn't ask that exactly 3 are heads, it asks the first three to be heads!
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 [#permalink] New post 18 May 2003, 15:52
I remember now that I had 1/8 as the answer in my mind when i framed the question.........it only mentions about the first three heads......so essentially the last two throws are useless.....

the answer will always be 1/8 irrespective of the number of throws.......i wonder why i did not reply to stolyar's posting........or maybe i did thru a PM. :wink:
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Re: nope [#permalink] New post 18 May 2003, 21:13
gravedigger wrote:
Stolyar, once again i disagree :roll:

p(First is heads) = 1/2
p(Second is heads) = 1/2
p(Third is heads) = 1/2

Events are independent, so p(ABC) = p(A)p(B)p(C)

p = 1/2 ^ 3 = 1/8!

In your case you forget that HHHHT and HHHTH also qualify as favorable. It doesn't ask that exactly 3 are heads, it asks the first three to be heads!


What can I say? Agree with your disagreement
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 [#permalink] New post 20 May 2003, 08:42
The correct logic works like this:

each toss has 2 outcomes. So 5 toss has total 2^5 = 32 oucomes.

The favourable outcomes are : HHHTT, HHHHT, HHHHH, HHHTH

So the required probability = 4/32 = 1/8.
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1,241,352 [#permalink] New post 20 May 2003, 09:58
your logic is correct, but...

Say, if instead of 5 coins we had 1,241,352 coins. What is the p that first 3 coins will have heads? It is still 1/8, but i doubt you can calculate 2^1,241,352! :lol:

Best is to use 1/2^3. Three independent events.

I guess this is what brstorewala means when saying the answer is
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1/8 irrespective of the number of throws.
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 [#permalink] New post 20 May 2003, 12:05
Hi gravedigger,
I like your logic :cool . It's really an independent thought :wink:

But i did like to say something when you say "i doubt you can calculate
2 ^1,241,352".

When, am1974 came out with cases, HHHTT,HHHTH... so on

He/She emphaised the point that there are 4 such cases. that means the last two places can be filled in 2^2 ways.

So going by the same way i know in a coin toss of 1,241,352. If first three are heads, the rest 1,241,349 can be filled in 2^1,241,349 cases.

So without calculating 2^1,241,352 :) i have the answer.

1/2^3.

The moral of the story....answer is 1/8 by either method with the minimal calculation. :D

:punk
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:) [#permalink] New post 21 May 2003, 03:01
:) great! it's a resourceful topic. You're right. THere're plenty ways of calculating it.
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my view [#permalink] New post 01 Jun 2003, 22:44
The answer is 4/32 = 1/8

2^5

and

HHHTT
HHHHH
HHHHT
HHHTH
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hey....I got 1/8 too!!! [#permalink] New post 06 Jun 2003, 12:00
Well,Here is how I woked it out.....
P(getting a head)=1/2 so P(getting a head first three times)=1/2*1/2*1/2=1/8!!!!!!!Ofcourse.the p(of getting a tail )= 1/2 too.I am not sure about this logic!!!! :stupid
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 [#permalink] New post 14 Mar 2006, 19:27
hmmm..neema 1/2*1/2*1/2 is the same raised to the power....
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 [#permalink] New post 14 Mar 2006, 21:37
The last two can be head or tail. # of ways for last two toss = 2*2 = 4

Probability = (1/2)^5 * 4 = 1/8
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 [#permalink] New post 16 Mar 2006, 09:25
HHHxy

total combinations of 3 Hs are 4.

4/(2^5) = 1/8
  [#permalink] 16 Mar 2006, 09:25
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