If you flip a coin twice, what is the probability of getting heads at least once

Complete step-by-step answer:
When a coin is tossed then, there are in total two possible outcomes i.e., HEAD and TAIL.
Let the total number of outcomes be denoted as n(T) and the total number of favorable outcomes be denoted as n(F).
Here in the question, the same coin has been tossed twice so, the total number of outcomes will be now 4 such that HEAD-HEAD, TAIL-TAIL, HEAD-TAIL and TAIL-HEAD. So, n(T)=4.
Now, we are interested in the probability of getting at least one head, which means that either one head or two heads but not zero heads. So, the favorable number of outcomes is 3 such as HEAD-HEAD, TAIL-HEAD and HEAD-TAIL. So, n(F)=3.
The ratio of the number of the favorable number of outcomes to the number of the total number of outcomes results in the probability of the event. Mathematically, $ P = \dfrac{{n(F)}}{{n(T)}} $
So, substituting the value of n(F) and n(T) in the equation $ P = \dfrac{{n(F)}}{{n(T)}} $ to determine the probability of getting atleast one head in tossing a coin twice.
 $
\Rightarrow P = \dfrac{{n(F)}}{{n(T)}} \\
   = \dfrac{3}{4} \\
  $
Hence, the probability of getting at least one head when the coin is tossed twice is $ \dfrac{3}{4} $ .

Note: Alternatively, we can also get the result by subtracting the probability of getting no heads with 1. So, the probability of getting at least one head = 1- probability of getting no heads.
 $
  P = 1 - \dfrac{1}{4} \\
   = \dfrac{3}{4} \\
  $

P(At least one head) = 1 – 0.5n

where:

• n: Total number of flips

For example, suppose we flip a coin 2 times.

The probability of getting at least one head during these 3 flips is:

• P(At least one head) = 1 – 0.5n
• P(At least one head) = 1 – 0.53
• P(At least one head) = 1 – 0.125
• P(At least one head) = 0.875

This answer makes sense if we list out every possible outcome for 2 coin flips with “T” representing tails and “H” representing heads:

• TTT
• TTH
• THH
• THT
• HHH
• HHT
• HTH
• HTT

Notice that at least one head (H) appears in 7 out of 8 possible outcomes, which is equal to 7/8 = 0.875.

Or suppose we flip a coin 5 times.

The probability of getting at least one head during these 5 flips is:

• P(At least one head) = 1 – 0.5n
• P(At least one head) = 1 – 0.55
• P(At least one head) = 1 – 0.25
• P(At least one head) = 0.96875

The following table shows the probability of getting at least one head during various amounts of coin flips:

Notice that the higher number of coin flips, the higher the probability of getting at least one head.

This should make sense considering the fact that we should have a higher probability of eventually seeing a head appear if we keep flipping the coin more times.

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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