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Probability of getting a head in coin flip is 1/2. If the coin is flipped two times what is the probability of getting a head in either of those attempts?

I think both the coin flips are mutually exclusive events, so the probability would be getting head in attempt 1 or attempt 2 which is :

P(attempt1) + P(attempt2) = 1/2 + 1/2 = 1

100% probability sounds wrong? What am I doing wrong. If i apply the same logic then probability of getting at least 1 head in 3 attempt will be 1/2+1/2+1/2 = 3/2 = 1.5 which I know for sure is wrong. What do i have mixed up?

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The events are not mutually exclusive: you can get a head on the first and on the second flip – Rookatu Mar 28 '14 at 5:55
Indeed. The coin tosses are independent. That is not the same as mutually exclusive. – Graham Kemp Mar 28 '14 at 9:58
up vote 6 down vote accepted

You are confusing the terms "independent" and "mutually exclusive".  These are not the same thing.  In fact events cannot be both "independent" and "mutually exclusive".  It's either one, the other, or neither.

"Mutually exclusive" simply means that the two events cannot happen together. If A happens then B does not and if B happens A cannot.

"Independent" simply means that the occurrence of one event is not conditional on the occurrence of the other. The probability of A happening does not depend on whether B happens or not, and vice versa.

Let $H_n$ be the indexed event of getting a head on the $n^{th}$ flip.

Given an unbiased coin, $P(H_1)=P(H_2)=\frac 1 2$

These events are independent so $P(H_1 \cap H_2) = P(H_1)\times P(H_2)$.  The outcome of one coin toss does not influence then outcome of the other.

However they are not mutually exclusive, so $P(H_1 \cup H_2) = P(H_1)+P(H_2) - P(H_1 \cap H_2)$.  Both coins can turn up heads.

Putting it together: $$\therefore P(H_1 \cup H_2) = \frac 12 + \frac 1 2 - \frac 12 \times \frac 12 = \frac 3 4$$

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So if I flip a coin 2 times, there is a 75% chance that I will get heads on at least one of the 2 flips? – JuanVelez Aug 25 '15 at 16:47
@JuanVelez Yes, if there is no bias (ie: fair coins). $\boxed{\begin{array}{c|cc} ~& H & T \\ \hline H & HH & HT \\ T & TH & TT\end{array}}$ Four equally probable results, three of which are the favoured outcome. – Graham Kemp Aug 25 '15 at 22:52
Would it follow, that in 3 tries you'd have an 87.5% chance of success? – Lonnie Best Dec 20 '15 at 23:15
@LonnieBest Yes, it does indeed follow that the change of at least one success in three flips is $1-0.5^3$. – Graham Kemp Dec 21 '15 at 0:29

Let $A$ be the event of getting a tail in both tosses, then $A'$ be the event of getting a head in either tosses. So $P(A') = 1 - P(A) = 1 - 0.5*0.5 = 0.75$

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probability of having head in a coin flip is 1/2 , when you flip 2 times then the probability you have at least 1 head is equal to :

1 - P(no head) = 1 - P(2 tail) = 1 - 1/2*1/2 = 3/4 or 75%

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I haven't seen the correct result yet. This one is easy as it is easy to run the possible results. They key is that if you get Heads the first time you don't need to flip again. So when people list 'HH or 'HT' they are wrong.

Options: H TH TT

2 out of 3 possibilities have 'H' = 2/3 = 66.66%

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Welcome to MathSE! Please be careful that your answer is accurate: this answer is not. – Rory Daulton Feb 26 at 1:27
The issue with your solution is that the Heads/No Flip situation will occur 50% of the time, not 33% of the time. – Keith Mar 19 at 17:53

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