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Q1(i):

Complete the following statements: (i) Probability of an event $E$ + Probability of the event ‘not $E$’ =              .

Solution :

Given: An event $E$ associated with a random experiment and its corresponding complementary event 'not $E$', denoted as $E^c$ or $\overline{E}$.

To Find: The sum of the probability of event $E$ and the probability of the event 'not $E$'.

Step 1: Defining the Probability of an Event
Let $S$ be the sample space of a random experiment. Let $E$ be an event such that $E \subseteq S$. The probability of event $E$, denoted by $P(E)$, is defined as the ratio of the number of favorable outcomes to the total number of equally likely outcomes in the sample space $S$.
$P(E) = \frac{\text{Number of outcomes favorable to } E}{\text{Total number of possible outcomes in } S}$

Step 2: Defining the Complementary Event
The event 'not $E$', denoted as $\overline{E}$, consists of all outcomes in the sample space $S$ that are not in $E$. Mathematically, $\overline{E} = S \setminus E$.
Since $E$ and $\overline{E}$ are mutually exclusive (they cannot occur simultaneously) and exhaustive (their union covers the entire sample space $S$), we have:
$E \cup \overline{E} = S$
$E \cap \overline{E} = \emptyset$

Step 3: Applying the Axiom of Probability
According to the axiomatic definition of probability, the sum of the probabilities of all elementary events in a sample space is equal to $1$.
Since $E$ and $\overline{E}$ partition the sample space $S$, the sum of their probabilities must equal the probability of the sure event (the sample space itself).
$P(E) + P(\overline{E}) = P(S)$

Step 4: Final Calculation
The probability of the sure event (the entire sample space $S$) is always $1$.
$P(S) = 1$
Therefore:
$P(E) + P(\text{not } E) = 1$

Final Answer: 1


More Questions from Class 10 Mathematics Probability EXERCISE 14.1


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