If P(E) = 1 then E is a
$
  (a) {\text{ impossible event}} \\
  (b) {\text{ sure event}} \\
  (c) {\text{ either event}} \\
  (d) {\text{ none of these}} \\
 $

Hint: The probability of any event always lies in the range $\left[ {0,1} \right]$, we can’t have probabilities greater than 1 and it can’t be even less than 0. Try to think of an event in which the probability is absolute 1, and then think from the options given that which suits best.

Complete step-by-step answer:

As we know that probability is the ratio of favorable number of outcomes to the total number of outcomes.
Let us consider an event E.
Therefore P(E) = $\dfrac{{{\text{Favorable number of outcomes}}}}{{{\text{Total number of outcomes}}}}$.
Now it is given that P(E) = 1.
Therefore P(E) = $\dfrac{{{\text{Favorable number of outcomes}}}}{{{\text{Total number of outcomes}}}} = 1$.
Therefore a favorable number of outcomes is equal to the total number of outcomes.
This is only possible if it is called a sure event or a certain event.
For example what is the probability of getting a number less than 7 when a dice is thrown.
As we know in a dice the possible outcomes are (1, 2, 3, 4, 5 and 6)
So the favorable number of outcomes is equal to the total number of outcomes.
Therefore P(E) = $\dfrac{6}{6} = 1$ which is a sure event or a certain event.
So this is the required answer.
Hence option (A) is correct.

Note: The range $\left[ {0,1} \right]$ has a different meaning from the range $\left( {0,1} \right)$. In prior o and 1 both are inclusive in the range however in the later one 0 and 1 are not included. If the probability of an event is 0 that means it’s an impossible event. Example what is the probability of obtaining a head when a dice is thrown. It is 0 because this can’t happen.