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
