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

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Manager
Joined: 28 Feb 2003
Posts: 100
Five coins are tossed one after the other. What is the [#permalink]

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24 Mar 2003, 22:11
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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.

don't me

bhavesh
SVP
Joined: 03 Feb 2003
Posts: 1604

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24 Mar 2003, 23:47
P(HHHTT)=5C3/32=10/32=5/16

Total outcomes: 2^5=32
CTO
Joined: 19 Dec 2002
Posts: 250
Location: Ukraine

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18 May 2003, 07:17
Stolyar, once again i disagree

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!
Manager
Joined: 28 Feb 2003
Posts: 100

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18 May 2003, 16:52
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.
SVP
Joined: 03 Feb 2003
Posts: 1604

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18 May 2003, 22:13
gravedigger wrote:
Stolyar, once again i disagree

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
Manager
Joined: 11 Mar 2003
Posts: 54
Location: Chicago

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20 May 2003, 09:42
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.
CTO
Joined: 19 Dec 2002
Posts: 250
Location: Ukraine

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20 May 2003, 10:58

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!

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

I guess this is what brstorewala means when saying the answer is
Quote:
1/8 irrespective of the number of throws.
Manager
Joined: 08 Apr 2003
Posts: 150

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20 May 2003, 13:05
Hi gravedigger,
I like your logic . It's really an independent thought

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

CTO
Joined: 19 Dec 2002
Posts: 250
Location: Ukraine

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21 May 2003, 04:01
great! it's a resourceful topic. You're right. THere're plenty ways of calculating it.
Intern
Joined: 25 May 2003
Posts: 17
Location: Thailand

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01 Jun 2003, 23:44
The answer is 4/32 = 1/8

2^5

and

HHHTT
HHHHH
HHHHT
HHHTH
Intern
Joined: 06 Jun 2003
Posts: 3

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06 Jun 2003, 13:00
Well,Here is how I woked it out.....
_________________

Its not about you getting knocked down,Its about how fast you get back up again!!!!

Intern
Joined: 18 Feb 2005
Posts: 47

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14 Mar 2006, 20:27
hmmm..neema 1/2*1/2*1/2 is the same raised to the power....
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5043
Location: Singapore

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14 Mar 2006, 22:37
The last two can be head or tail. # of ways for last two toss = 2*2 = 4

Probability = (1/2)^5 * 4 = 1/8
SVP
Joined: 14 Dec 2004
Posts: 1689

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16 Mar 2006, 10:25
HHHxy

total combinations of 3 Hs are 4.

4/(2^5) = 1/8
16 Mar 2006, 10:25
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