## Probability - conceptual problem

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### Probability - conceptual problem

by vikram4689 » Tue Aug 07, 2012 4:31 pm
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
= 1/2*1
P = 1/2
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by [email protected] » Tue Aug 07, 2012 4:44 pm
vikram4689 wrote:what is the probability of getting at least one head in a 2 times toss of coin.

i know how to calculate this but what is CONCEPTUAL PROBLEM in following solution
= 1/2*1
P = 1/2
Your solution assumes that the first toss will be heads. This need not be the case in order to get at least one head.

When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(getting at least 1 head) = 1 - P(not getting at least 1 head)
What does it mean to not get at least 1 head? It means getting zero heads.
So, we can write: P(getting at least 1 head) = 1 - P(getting zero heads)

Now let's calculate P(getting zero heads)
What needs to happen in order to get zero heads?
Well, we need tails on the first toss and tails on the second toss.
We can write P(getting zero heads) = P(tails on 1st AND tails on 2nd)
This means that P(getting zero heads) = P(tails on 1st) x P(tails on 2nd)
Which means P(getting zero heads) = (1/2)x(1/2) = 1/4

P(getting at least 1 head) = 1 - P(not getting at least 1 head)
= 1 - 1/4
= 3/4

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Brent

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by [email protected] » Tue Aug 07, 2012 4:48 pm
Another way to handle this question is to examine all of the possible outcomes when we toss a coin two times. The outcomes are as follows:
- heads on 1st and tails on 2nd
- tails on 1st and heads on 2nd
- tails on 1st and tails on 2nd

Important: Please note that these 4 outcomes are equally likely to occur.

Out of the 4 possible (and equally likely) outcomes, 3 of them have at least one heads.
So, P(at least one heads) = 3/4

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by vikram4689 » Tue Aug 07, 2012 5:34 pm
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.
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by [email protected] » Tue Aug 07, 2012 5:40 pm
I guess it's unclear to me what you meant by P = 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.

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by vikram4689 » Tue Aug 07, 2012 5:45 pm
[email protected] wrote:I guess it's unclear to me what you meant by P = 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

please let me know if my explanation is not clear
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by eagleeye » Tue Aug 07, 2012 6:11 pm
vikram4689 wrote:
[email protected] wrote:I guess it's unclear to me what you meant by P = 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

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

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by lunarpower » Thu Aug 16, 2012 11:43 pm
whoa, there are, um, a lot of words in this thread.

vikram, the basic issue with your approach is this: when you multiply the probabilities of consecutive events, that multiplication implies that ORDER MATTERS.

so, when you write "prob(head) x prob(anything)", you are writing an expression for the probability that the first toss is heads. unsurprisingly, that probability works out to 1/2.

because order always matters when you multiply probabilities, it's rather annoying to get your approach to work here. specifically, you'd have to do the following:
this is your standard probability calculation for either of two events. in this case, it's prob(heads on first coin) + prob(heads on second coin) - prob(both).
this works out to 1/2 + 1/2 - 1/4 = 3/4, as required.
Ron has been teaching various standardized tests for 20 years.

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