A coin was tossed three times what is the probability that at least 1 tail will occur

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What is the probability that at least 3 coin tosses will be needed?

I am at a loss on how to calculate this probability. I have found that the sample space will be

$S=2^n-2$

For two tosses, 2 out of 4 will follow the condition. For three tosses, 6 out of 8 will follow the condition. For four tosses, 14 out of 16 will follow the condition. And so on. Reading a bit about how to go about the solution. I have found that the expected number of coin flips for at least 1 head and at least 1 tail will be 3 (still kind of unsure as to how this is true). https://www.quora.com/What-is-the-expected-number-of-coin-flips-until-you-get-one-head-and-one-tail

Is this on the right path on how to find the probability?

It is worth going through the effort of calculating the probability via the definition of conditional probability in early examples.

$$Pr(A\mid B):=\frac{Pr(A\cap B)}{Pr(B)}$$

Let $B$ be the event that the first coin flipped is not a head (i.e. the first coin flipped turned up tails).

Let $A$ be the event that the coin is flipped exactly three times.

We are tasked with calculating $Pr(A\mid B)$, the probability that the coin is flipped exactly three times given that the first flip did not turn up heads.

We can draw ourselves a tree diagram or however else we like to arrive at the following table of outcomes and respective probabilities:

$$\begin{array}{|c|c|}\hline\text{Outcome}&\text{Probability}\\\hline H&\frac{1}{2}\\\hline TH&\frac{1}{4}\\\hline TTH&\frac{1}{8}\\\hline TTT&\frac{1}{8}\\\hline\end{array}$$

It is worth taking a moment to check that this does in fact make sense as a probability distribution by verifying that the probabilities add up to exactly one. Indeed $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}$ does equal $1$.

The event that the first flip is not heads corresponds to all of the above listed outcomes except the first and so occurs with probability $\frac{1}{4}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2}$ so we learn that $Pr(B)=\frac{1}{2}$.

The event that the first flip is not heads and it takes three flips in total corresponds to the last two outcomes in the above table and so occurs with probability $\frac{1}{8}+\frac{1}{8}=\frac{1}{4}$ so we learn that $Pr(A\cap B)=\frac{1}{4}$.

Putting this information together, we get:

$$Pr(A\mid B)=\frac{Pr(A\cap B)}{Pr(B)}=\frac{1/4}{1/2}=\frac{1}{2}$$

To find the probability of getting at least one tail in Tossing a coin three time,let the event is A

we know that Probability = Favourable outcome/total outcome

total outcomes:8

these are

[TTT,TTH,THT,THH,HTT,HTH,HHT,HHH}

Favourable outcome of getting at least one tail:7 these are :{TTT,TTH,THT,THH,HTT,HTH,HHT}

Probability of getting at least one tail

p(E) =7/8

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We will be using the concept of probability to solve this.

Answer: The probability of flipping a coin three times and getting 3 tails is 1/8.

Let's solve this step by step.

Explanation:

Let us mark H for Heads and T for Tails.

The coin is flipped three times; the total number of outcomes = 2 × 2 × 2 = 8.

Sample Space (|Ω|) = {(H, H, H), (H, H, T), (H, T, H), (H, T, T), (T, H, H), (T, H, T), (T, T, H), (T, T, T)} = 8

Let X be the probability of getting all tails: X = {(T, T, T)}

⇒ |X| = 1

P(X) = |X| / |Ω| = 1/8.

Thus, the probability of flipping a coin three times and getting 3 tails is 1/8.

getcalc.com's solved example with solution to find what is the probability of getting 1 Tail in 3 coin tosses.
P(A) = 7/8 = 0.88 for total possible combinations for sample space S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} & successful events for getting at least 1 tail A = {HHT, HTH, HTT, THH, THT, TTH, TTT} for an experiment consists of three independent events.

for 1 Tail in 3 Coin Flips
	Atleast 1 Tail	Exactly 1 Tail
Total Events n(S)	8	8
Success Events n(A)	7	3
Probability P(A)	0.88	0.38

The above probability of outcomes applicable to the below questions too.

Probability of flipping a coin 1 times and getting 3 tail in a row
Probability of getting 3 tail when flipping 1 coins together
A coin is tossed 1 times, find the probability that at least 3 are tail?
If you flip a fair coin 1 times what is the probability that you will get exactly 3 tail?
A coin is tossed 1 times, what is the probability of getting exactly 3 tail?

The ratio of successful events A = 7 to the total number of possible combinations of a sample space S = 8 is the probability of 1 tail in 3 coin tosses. Users may refer the below solved example work with steps to learn how to find what is the probability of getting at-least 1 tail, if a coin is tossed three times or 3 coins tossed together. Users may refer this tree diagram to learn how to find all the possible combinations of sample space for flipping a coin one, two, three or four times.

Solution

Step by step workout
step 1 Find the total possible events of sample space S S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} S = 8

step 2 Find the expected or successful events A

A = {HHT, HTH, HTT, THH, THT, TTH, TTT} A = 7

step 3 Find the probability

P(A) = Successful Events/Total Events of Sample Space
= 7/8 = 0.88 P(A) = 0.88

0.88 is the probability of getting 1 Tail in 3 tosses.

The ratio of successful events A = 3 to total number of possible combinations of sample space S = 8 is the probability of 1 tail in 3 coin tosses. Users may refer the below detailed solved example with step by step calculation to learn how to find what is the probability of getting exactly 1 tail, if a coin is tossed three times or 3 coins tossed together.

Solution :

Step by step workout
step 1 Find the total possible combinations of sample space S S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} S = 8

step 2 Find the expected or successful events A

A = {HHT, HTH, THH} A = 3

step 3 Find the probability

P(A) = Successful Events/Total Events of Sample Space
= 3/8 = 0.38 P(A) = 0.38

0.38 is the probability of getting exactly 1 Tail in 3 tosses.