[Blindmath] Seemingly basic math question involving probability

[Ramana Polavarapu]

Yes.  You are dealing with the following 8 combinations:
Hhh
Hht
Hth
Htt
Thh
Tht
Tth
Ttt
All of them have the same probability (1/8) as long as the coin is fair. 

-----Original Message-----
From: blindmath-bounces at nfbnet.org [mailto:blindmath-bounces at nfbnet.org] On
Behalf Of Matthew_2010
Sent: Thursday, October 01, 2009 11:33 AM
To: Blind Math list for those interested in mathematics
Subject: Re: [Blindmath] Seemingly basic math question involving probability

Thanks for your reply. I'm still a little baffled as to the significance of
the tails in flip #3. Ultimately, it appears as though the fact that the
third flip should be tales is insignificant in the grand scheme of this
probability question, right?
H H T = 1/8
HTH = 1/8
THH = 1/8
I guess the tales part only served to confused me a bit.

Thanks,

Matthew

----- Original Message -----
From: "Ramana Polavarapu" <sriramana at gmail.com>
To: "'Blind Math list for those interested in mathematics'" 
<blindmath at nfbnet.org>
Sent: Wednesday, September 30, 2009 9:13 PM
Subject: Re: [Blindmath] Seemingly basic math question involving probability


> As far as I can see, you are on the right track.  However, the probability
> should be 1/2 but not 1/3.  We apply the multiplicative rule since we want
> them to appear in the order specified.  Therefore, the right answer is 
> 1/8.
>
> Regards,
>
> Ramana
>
>
> -----Original Message-----
> From: blindmath-bounces at nfbnet.org [mailto:blindmath-bounces at nfbnet.org] 
> On
> Behalf Of Matthew_2010
> Sent: Thursday, October 01, 2009 4:26 AM
> To: Blind Math list for those interested in mathematics
> Subject: [Blindmath] Seemingly basic math question involving probability
>
> Alright all, I'm embarrassed to ask but here goes.
>
> Yesterday one of my kids asked me to help with a math problem he couldn't
> work out and its been driving me a little crazy, so here I am. He was 
> asked
> to determine the probability of flipping a coin and getting heads on first
> flip, heads on second flip, and tails on third flip. I'm thinking that
> because these are independent events its a matter of multiplying 1/3 * 1/3

> *
> 1/3. Therefore, the answer is that there is a 4% chance of flipping 3 
> times
> and getting this pattern.
>
> The part of the problem that's bugging me is the tails part. Shouldn't 
> this
> be a different probability since its the third flip and specified as 
> tails?
> In other words, if the third flip was heads it wouldn't change the
> probability at all since they would all be equal (1/3). What role does a
> third tails flip play in the matter? (Hmm).
>
> Matthew
>
>
>
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