
The probability of an event lies between:
(a)\[-1\le P\le 1\]
(b) \[0\le P\le 1\]
(c) \[-1\le P\le 0\]
(d) None of these.
Answer
623.4k+ views
- Hint: In this question, we first need to get the relation between the number of favourable outcomes and the total number of outcomes possible. Then we can get the inequality for the probability.
Complete step-by-step solution -
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
\[P\left( A \right)=\dfrac{m}{n}\]
That means probability of any event is the ratio of number of favourable outcomes to the total number of possible outcomes.
Now, as we already know that the number of favourable outcomes are the possibilities for the given condition which is a subset of the total number of possible outcomes.
Since the number of elements in the set of favourable outcomes will always be less than or equal to the total number of possible outcomes.
So, the probability of any event cannot be more than 1 which means that it will always be less than 1 that implies the maximum value of probability of an event is 1.
Now, as the number of favourable outcomes is the possibility of outcomes for the given condition it cannot be negative but can be 0 or greater than that.
Thus, the probability of any event can be greater than or equal to 0 and less than or equal to 1.
\[0\le P\le 1\]
Hence, the correct option is (b).
Note: Let us consider an example of throwing a die in which the possibilities for the outcome are
\[\left\{ 1,2,3,4,5,6 \right\}\]. Now, the probability for getting a number less than 1 will be 0 as there are no favourable outcomes for that. Then the probability of getting a number greater than 3 is \[\dfrac{1}{2}\] as the favourable outcomes are 3 and the total possible outcomes are 6.
Now, the probability of getting a number greater than 0 is 1 because the total number of possible outcomes are 6 and the number of favourable outcomes for the given condition are also 6.
Complete step-by-step solution -
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
\[P\left( A \right)=\dfrac{m}{n}\]
That means probability of any event is the ratio of number of favourable outcomes to the total number of possible outcomes.
Now, as we already know that the number of favourable outcomes are the possibilities for the given condition which is a subset of the total number of possible outcomes.
Since the number of elements in the set of favourable outcomes will always be less than or equal to the total number of possible outcomes.
So, the probability of any event cannot be more than 1 which means that it will always be less than 1 that implies the maximum value of probability of an event is 1.
Now, as the number of favourable outcomes is the possibility of outcomes for the given condition it cannot be negative but can be 0 or greater than that.
Thus, the probability of any event can be greater than or equal to 0 and less than or equal to 1.
\[0\le P\le 1\]
Hence, the correct option is (b).
Note: Let us consider an example of throwing a die in which the possibilities for the outcome are
\[\left\{ 1,2,3,4,5,6 \right\}\]. Now, the probability for getting a number less than 1 will be 0 as there are no favourable outcomes for that. Then the probability of getting a number greater than 3 is \[\dfrac{1}{2}\] as the favourable outcomes are 3 and the total possible outcomes are 6.
Now, the probability of getting a number greater than 0 is 1 because the total number of possible outcomes are 6 and the number of favourable outcomes for the given condition are also 6.
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