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5 coins are tossed. If two of them show heads, then the

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5 coins are tossed. If two of them show heads, then the [#permalink] New post 31 Mar 2005, 17:36
5 coins are tossed. If two of them show heads, then the probability that all 5 coins show head is

a. 1/32
b. 1/10
c. 1/26
d. 1/13
e. none of the above
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 [#permalink] New post 31 Mar 2005, 18:38
the 2 coins prob is 1

rest 3 coins prob is (1/2)^3 = 1/8
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 [#permalink] New post 31 Mar 2005, 18:43
OA on this is C. (1/26)
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 [#permalink] New post 31 Mar 2005, 21:54
I got A...what am i doing wrong agian
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 [#permalink] New post 31 Mar 2005, 22:05
Total number of 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 # of combinations = 32 ways

Number of combinations that all are heads, is only 1: HHHHH

So Probability = 1/32
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Re: coin [#permalink] New post 01 Apr 2005, 09:06
cloaked_vessel wrote:
5 coins are tossed. If two of them show heads, then the probability that all 5 coins show head is

a. 1/32
b. 1/10
c. 1/26
d. 1/13
e. none of the above


Total outcome:2^5=32
Outcome of at least two of them show heads:
Total-(1 head)-(0 head)=32-5-1=26
Outcome of 5 heads=1
Probability=1/26
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Last edited by HongHu on 01 Apr 2005, 09:50, edited 1 time in total.
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Re: coin [#permalink] New post 01 Apr 2005, 09:38
HongHu wrote:
cloaked_vessel wrote:
5 coins are tossed. If two of them show heads, then the probability that all 5 coins show head is

a. 1/32
b. 1/10
c. 1/26
d. 1/13
e. none of the above


Total outcome:2^5=32
Outcome of at least two of them show heads:
Total-(1 head)-(0 head)=32-1-5=26
Outcome of 5 heads=1
Probability=1/26


why do we have to subtract 5 from 32.
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 [#permalink] New post 01 Apr 2005, 09:50
Sorry, the 1 is for 0 head, and the 5 is for 1 head. I'll edit the post to make it clearer.

Basically the question implies that two of them are heads but the other three can also be heads. So we need to know what probability is for getting at least two heads. It'll be total minus the cases where we only get one head, and where we get no heads.
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Re: coin [#permalink] New post 01 Apr 2005, 11:04
HongHu wrote:
cloaked_vessel wrote:
5 coins are tossed. If two of them show heads, then the probability that all 5 coins show head is

a. 1/32
b. 1/10
c. 1/26
d. 1/13
e. none of the above


Total outcome:2^5=32
Outcome of at least two of them show heads:
Total-(1 head)-(0 head)=32-5-1=26
Outcome of 5 heads=1
Probability=1/26


Honghu, how do u know that "If two of them show heads" means "atleast 2 heads"...thx
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 [#permalink] New post 01 Apr 2005, 12:19
There are two explanations for this sentence "two of the coins show heads". One is that exactly two coins are heads. The other is at least two are heads. The next sentence askes "What probability is all coins show heads". You know that means in additional to the two heads, the other coins may show heads too. In other word there would be two or more heads, or at least two heads.
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 [#permalink] New post 02 Apr 2005, 03:26
Sorry guys but I don't get this one. the question is :

5 coins are tossed. If two of them show heads, then the probability that all 5 coins show head is ?

We already know that 2 of them showed heads, no ? In my opnion it should be 1/8 becasue there are still 3 toss to make and the events are independant : (1/2)^3

What's wrong with 1/8 ?
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 [#permalink] New post 04 Apr 2005, 08:01
Because it didn't say that the first two throw were heads. It just said that two out of five are heads and we don't know the others.
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 [#permalink] New post 06 Apr 2005, 04:04
Thanks Hong Hu, I got it now
  [#permalink] 06 Apr 2005, 04:04
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