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A coin is tossed twice, what is the probability that at least one tail occurs?
Answer
511.2k+ views
Hint: In this question, we first need to list all the possibilities when two coins are tossed. Then check the favourable outcomes and substitute them in the probability formula to get the result
\[P\left(A \right)=\dfrac{m}{n}=\dfrac{\text{number of favourable outcomes}}{\text{total number of possible outcomes}}\].
Complete step-by-step answer:
SAMPLE SPACE: The set of all possible outcomes of an experiment is called the sample space of the experiment and is denoted by S.
PROBABILITY: If there are n elementary events associated with a random experiment and m of them are favourable to an event A, then the probability of happening or occurrence of A, denoted by P(A), is given by
\[P\left( A \right)=\dfrac{m}{n}=\dfrac{\text{number of favourable outcomes}}{\text{total number of possible outcomes}}\]
Given in the question that 2 coins are being tossed.
Let us now write the sample space for tossing of two coins.
Let us assume the representation of tail as T and head as H.
\[S=\left\{ H,H \right\},\left\{ H,T \right\},\left\{ T,H \right\},\left\{ T,T \right\}\]
Now, we have the total number of possible outcomes listed in the sample space.
Let us assume that the event for getting at-least one tail as A.
Now, to get at-least one tail when we toss two coins means that either one of the coin should show up tails or both the coins may show up tails.
Here, the number of favourable outcomes to get at-least one tail are 3 and the total number of possible outcomes when two coins are tossed are 4.
Now, on comparing these values with the probability formula we get,
\[m=3,n=4\]
Now, by substituting these values in the probability formula we get,
\[\begin{align}
& \Rightarrow P\left( A \right)=\dfrac{m}{n} \\
& \therefore P\left( A \right)=\dfrac{3}{4} \\
\end{align}\]
Note: It is important to note that while considering the number of favourable outcomes we also need to consider the case in which both the coins show up tails because at-least one tail means either it can be one or more than one. While substituting the values respective values should be substituted because changing the numerator and denominator changes the whole result.
\[P\left(A \right)=\dfrac{m}{n}=\dfrac{\text{number of favourable outcomes}}{\text{total number of possible outcomes}}\].
Complete step-by-step answer:
SAMPLE SPACE: The set of all possible outcomes of an experiment is called the sample space of the experiment and is denoted by S.
PROBABILITY: If there are n elementary events associated with a random experiment and m of them are favourable to an event A, then the probability of happening or occurrence of A, denoted by P(A), is given by
\[P\left( A \right)=\dfrac{m}{n}=\dfrac{\text{number of favourable outcomes}}{\text{total number of possible outcomes}}\]
Given in the question that 2 coins are being tossed.
Let us now write the sample space for tossing of two coins.
Let us assume the representation of tail as T and head as H.
\[S=\left\{ H,H \right\},\left\{ H,T \right\},\left\{ T,H \right\},\left\{ T,T \right\}\]
Now, we have the total number of possible outcomes listed in the sample space.
Let us assume that the event for getting at-least one tail as A.
Now, to get at-least one tail when we toss two coins means that either one of the coin should show up tails or both the coins may show up tails.
Here, the number of favourable outcomes to get at-least one tail are 3 and the total number of possible outcomes when two coins are tossed are 4.
Now, on comparing these values with the probability formula we get,
\[m=3,n=4\]
Now, by substituting these values in the probability formula we get,
\[\begin{align}
& \Rightarrow P\left( A \right)=\dfrac{m}{n} \\
& \therefore P\left( A \right)=\dfrac{3}{4} \\
\end{align}\]
Note: It is important to note that while considering the number of favourable outcomes we also need to consider the case in which both the coins show up tails because at-least one tail means either it can be one or more than one. While substituting the values respective values should be substituted because changing the numerator and denominator changes the whole result.
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