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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:

$$P(\text{attempt $1$}) + P(\text{attempt $2$}) = 1/2 + 1/2 = 1$$

$100\%$ 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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  • $\begingroup$ The events are not mutually exclusive: you can get a head on the first and on the second flip $\endgroup$
    – Rookatu
    Mar 28, 2014 at 5:55
  • $\begingroup$ Indeed. The coin tosses are independent. That is not the same as mutually exclusive. $\endgroup$
    – Graham Kemp
    Mar 28, 2014 at 9:58

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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 does not.

"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 the 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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    $\begingroup$ 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? $\endgroup$
    – Juan Velez
    Aug 25, 2015 at 16:47
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    $\begingroup$ @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. $\endgroup$
    – Graham Kemp
    Aug 25, 2015 at 22:52
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    $\begingroup$ Would it follow, that in 3 tries you'd have an 87.5% chance of success? $\endgroup$
    – Lonnie Best
    Dec 20, 2015 at 23:15
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    $\begingroup$ @LonnieBest Yes, it does indeed follow that the change of at least one success in three flips is $1-0.5^3$. $\endgroup$
    – Graham Kemp
    Dec 21, 2015 at 0:29
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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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After flipping the coin two times there are four possible outcomes: HH, HT, TH, TT.

Three of them have heads: HH, HT, TH.

So the probability of getting heads in two coin flipps is $3/4$.

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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:

$1 - P(\text{no head}) = 1 - P(2\, \text{tail}) = 1 - 1/2\times1/2 = 3/4$ 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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  • $\begingroup$ Welcome to MathSE! Please be careful that your answer is accurate: this answer is not. $\endgroup$
    – Rory Daulton
    Feb 26, 2016 at 1:27
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    $\begingroup$ The issue with your solution is that the Heads/No Flip situation will occur 50% of the time, not 33% of the time. $\endgroup$
    – Keith
    Mar 19, 2016 at 17:53
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    $\begingroup$ The question asks for the probability of tossing a head in two tosses, so HH very much applies (since you have tossed a head in two tries; the fact that you have tossed one H twice is neither here nor there, the thing that matters is that in two tosses the outcome HH produces a head.) $\endgroup$
    – Daniel Buck
    Jul 30, 2018 at 1:34

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