Problem Solving

what is the probability of getting at least one head when a fair coin is flipped twice ?

i know how to calculate this (ans=3/4) but what is CONCEPTUAL PROBLEM in following solution
P = P(Head)*P(Anything head or tail)
= 1/2*1
P = 1/2

brent,
i am aware of both of these solutions and concur with you on their correctness. however, i am not able to reason out the logical fallacy in the solution i posted

also i don't understand how did i assume that first toss will be head. simple example to check this. probability of head on single toss of coin => P(head)=1/2, if i did assume that head is the only possible then i would write P(head)=1

Your solution assumes that the first toss will be heads. This need not be the case in order to get at least one head.

Brent@GMATPrepNow wrote:I guess it's unclear to me what you meant by P = P(Head)*P(Anything head or tail)
Why do you feel that P(at least one head) = P(Head)*P(Anything head or tail)?
I guess if we can better understand your rationale for this equation, we can get to the heart of the conceptual problem.

Cheers,
Brent

sure,
since question asks for a least one head, solution says i need one head and i do not care what 2nd outcome is. also both of these have to happen for the event to be complete, we AND individual probabilities
p(head)=1/2
p(anything head or tail)= p(head)+(tail)-p(head and tail)= 1/2+1/2-0=1
hence,p = p(Head)*p(Anything head or tail)=1/2*1=1/2

please let me know if my explanation is not clear

vikram4689 wrote:

Brent@GMATPrepNow wrote:I guess it's unclear to me what you meant by P = P(Head)*P(Anything head or tail)
Why do you feel that P(at least one head) = P(Head)*P(Anything head or tail)?
I guess if we can better understand your rationale for this equation, we can get to the heart of the conceptual problem.

Cheers,
Brent

sure,
since question asks for a least one head, solution says i need one head and i do not care what 2nd outcome is. also both of these have to happen for the event to be complete, we AND individual probabilities
p(head)=1/2
p(anything head or tail)= p(head)+(tail)-p(head and tail)= 1/2+1/2-0=1
hence,p = p(Head)*p(Anything head or tail)=1/2*1=1/2

please let me know if my explanation is not clear

Hi vikram4689:
I see the fallacy in your argument. The problem is not completely with the thought process but with the application.

First: When you are calculating probability in : P = (p head)* p(tail) etc., you are inadvertently and inherently considering the order of the outcomes. That's just the way it is.

So, when you say that you have P(head) = P(head)*P(head or tail) which you correctly calculated as 1/2*1 = 1/2 , you are only considering the cases when there is HH and HT.
A quick fix would be considering the other possibility when you have P(head or tail)*P(head). In that case, your probability would be 1/2 (from the first case) + 1/2 (from the 2nd case) and equal 1. You know intuitively that this is wrong.
Here's the other problem:

If you did it this way, you have counted the HH case twice, if you were to remove it from here, you would get the desired result. P(HH) = 1/2*1/2 = 1/4.

Hence required probability = (Head first) + (Head Second) - (Both heads) = 1/2 + 1/2 - 1/4.

Also, let's translate what you calculated back to words:
When you said P(required) = P(head)*P(anything), you are answering the following question:
If a fair coin is tossed twice, what is the probability that it always lands heads on the first turn?

Let me know if the fallacy is clear.

Someone else had a similar problem, and I posted an explanation here regarding what are we really calculating when we say P(required) = P(A) * P(B) etc :

https://www.beatthegmat.com/red-and-whit ... tml#490788

Let me know if this helps