The probability of an event lies between
[a] 1 and 1
[b] 1 and 0
[c] 0 and 1
[d] None of these.
Answer
643.2k+ views
Hint: Recall the definition of the probability of an event E of a random experiment X with sample space S. Use the fact that $\forall E\subset S,0\le n\left( E \right)\le n\left( S \right)$ and hence find the range of the probability of experiment E.
Complete step-by-step answer:
Before solving the question, we need to know what probability of an event means.
Consider a random experiment X and let S be the set of all possibilities of X. The set S is called the sample space of the random experiment of X. Consider any subset E of S. The set E is called an event.
The probability of an event E is defined as the ratio of the number of elements in E to the number of elements in S, i.e. $P\left( E \right)=\dfrac{\text{number of favourable cases}}{\text{Total number of cases}}=\dfrac{n\left( E \right)}{n\left( S \right)}$.
Now we know that $\forall E\subset S$, we have $0\le n\left( E \right)\le n\left( S \right)$
Dividing both sides by n(S), we get
$\dfrac{0}{n\left( S \right)}\le \dfrac{n\left( E \right)}{n\left( S \right)}\le \dfrac{n\left( S \right)}{n\left( S \right)}\Rightarrow 0\le P\left( E \right)\le 1$
Hence the probability of an event E lies between 0 and 1.
Hence option [c] is correct.
Note: [1] Although the above result has been proved only for countable sets E and S, it holds true for all sets E and S over which the probability is defined. The definition of the probability is beyond the scope of the current grade.
[2] The above result is used to check whether the given probability of an event is possible or not.
Complete step-by-step answer:
Before solving the question, we need to know what probability of an event means.
Consider a random experiment X and let S be the set of all possibilities of X. The set S is called the sample space of the random experiment of X. Consider any subset E of S. The set E is called an event.
The probability of an event E is defined as the ratio of the number of elements in E to the number of elements in S, i.e. $P\left( E \right)=\dfrac{\text{number of favourable cases}}{\text{Total number of cases}}=\dfrac{n\left( E \right)}{n\left( S \right)}$.
Now we know that $\forall E\subset S$, we have $0\le n\left( E \right)\le n\left( S \right)$
Dividing both sides by n(S), we get
$\dfrac{0}{n\left( S \right)}\le \dfrac{n\left( E \right)}{n\left( S \right)}\le \dfrac{n\left( S \right)}{n\left( S \right)}\Rightarrow 0\le P\left( E \right)\le 1$
Hence the probability of an event E lies between 0 and 1.
Hence option [c] is correct.
Note: [1] Although the above result has been proved only for countable sets E and S, it holds true for all sets E and S over which the probability is defined. The definition of the probability is beyond the scope of the current grade.
[2] The above result is used to check whether the given probability of an event is possible or not.
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