A coin is tossed 3 times, what is the probability of getting exactly one head

Last updated date: 03rd Jan 2023

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Answer

Hint:
Collect all the possible combinations for sample space. From that we get the total number of possible outcomes. Then we get the probability by using formula.

Complete step by step solution:
The coin is tossed three times the possible outcome number is $2\times 2\times 2=8$. The number two is the possibility of heads and tails when it is tossed three times we need to multiply by three times. Now each individual coin is equally likely to come up heads or tails then we have possible combinations are HHH, HHT, HTT, HTH, TTT, TTH, THH and THT. These are the sample space, sample space is nothing but a set of all possible outcomes of the experiment or results of that experiment. We denoted sample space in set notation, it is represented by the symbol “S''.
So we can write sample space in this format,
S= {HHH, HHT, HTT, HTH, TTT, TTH, THH, THT}
To find the probability we need to divide the favorable outcomes by the number of outcomes. A favorable outcome means that the number of events we want from the experiment.
$\Rightarrow $ Probability of A = favorable outcomes ÷number of outcomes
1) A: getting at least two heads,
Here we need the probability of getting at least two heads, at least two heads means getting two heads and more than that. To determine the results we need to check from the sample space which gives at least two heads.
HHH-yes, HHT-yes, HTT-no, HTH-yes, TTT-no, TTH-no, THH-yes, THT-no.
$\Rightarrow $ Here the favorable outcomes are HHT, HTH, THH and HHH.
$\Rightarrow $ Number of favorable outcomes is $4$
$\Rightarrow $ Number of outcomes here is $8$
$\Rightarrow P(A)=\dfrac{4}{8}$
$\Rightarrow P(A)=\dfrac{1}{2}$
Probability of getting at least two heads is $\dfrac{1}{2}$
2) B: getting exactly two heads,
Here we need to find the probability of getting exactly two heads means we have to get absolutely two heads. To determine the results we have to check from the sample space,
HHH-no, HHT-yes, HTT-no, HTH-yes, TTT-no, TTH-no, THH-yes, THT-no.
$\Rightarrow $ Here the favorable outcomes are HHT, HTH and THH.
$\Rightarrow $ Number of favorable outcomes is $3$
So here we have the probability as,
$\Rightarrow P(B)=\dfrac{3}{8}$
Probability of getting exactly two heads is $\dfrac{3}{8}$
3) C: getting at most one head,
Here we need to find the probability of getting at most one head.
At most one head means that only one of the three coins show up ahead or no head. To determine the results we have to check from the sample space,
HHH-no, HHT-no, HTT-yes, HTH-no, TTT-yes, TTH-yes, THH-no, THT-yes.
$\Rightarrow $ Here the favorable outcomes are HTT, TTT, TTH and THT.
$\Rightarrow $ Number of favorable outcomes is $4$
So apply the formula we get,
$\Rightarrow P(C)=\dfrac{4}{8}$
We can also write,
$\Rightarrow P(C)=\dfrac{1}{2}$
Probability of getting at most one head is $\dfrac{1}{2}$
4) D: getting more heads than tails,
Here we need to find the probability of getting more heads than tails.
This means the heads come up is more than tails come up. To determine the results we have to check from the sample space,
HHH-yes, HHT-yes, HTT-no, HTH-yes, TTT-no, TTH-no, THH-yes, THT-no.
$\Rightarrow $ Here the favorable outcomes are HHH, HHT, HTH and THH.
$\Rightarrow $ Number of favorable outcomes is $4$
So we get,
$\Rightarrow P(D)=\dfrac{4}{8}$
Then we have,
$\Rightarrow P(D)=\dfrac{1}{2}$
Probability of getting more heads than tails is $\dfrac{1}{2}$

Note:
A random experiment is an experiment in which
1) The set of all possible outcomes are known.
2) Exact outcome is not known.

1. \(\dfrac 1 4\)
2. \(\dfrac1 8\)
3. \(\dfrac 1 2\)
4. \(\dfrac 3 8\)

Answer (Detailed Solution Below)

Option 1 : \(\dfrac 1 4\)

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Electric charges and coulomb's law (Basic)

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Concept:

The probability of the occurrence of an event A, out of total possible outcomes N, is given by P(A) = \(\rm \dfrac{n(A)}{N}\), where n(A) is the number of ways in which the event A can occur.

Calculation:

The total number of different possible outcomes (N) in tossing a coin 3 times is 23 = 8.

For getting a head and a tail alternately, the possibilities are HTH, THT → 2 possibilities n(A).

∴ Required probability = \(\rm \dfrac{n(A)}{N}=\dfrac{2}{8}=\dfrac{1}{4}\)

Alternate Method

The possible set of A coin is tossed 3 times is {HHH}{HHT}{HTH}{HTT}{TTT}{TTH}{THT}{THH} = 8

The probability of getting a head and a tail alternately is {HTH}{THT} = 2

So, required probability = \(\rm \dfrac{n(A)}{N}=\dfrac{2}{8}=\dfrac{1}{4}\)

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What is the probability of getting one head when tossing a coin 3 times?

What is the probability of getting exactly 1 head?