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Q13(ii):
A die is thrown once. Find the probability of getting (ii) a number lying between 2 and 6;

Solution :

Given: A fair, standard six-faced die is thrown once. The possible outcomes on the faces of the die are $\{1, 2, 3, 4, 5, 6\}$.

To find: The probability of getting a number lying between 2 and 6.

Visual Representation of the Sample Space:

1 2 3 4 5 6

Step 1: Define the Sample Space ($S$)
The sample space $S$ consists of all possible outcomes when a die is thrown:
$S = \{1, 2, 3, 4, 5, 6\}$
The total number of possible outcomes, denoted by $n(S)$, is $6$.

Step 2: Define the Event ($E$)
Let $E$ be the event of getting a number lying between 2 and 6. The numbers strictly between 2 and 6 are 3, 4, and 5.
$E = \{3, 4, 5\}$
The number of favorable outcomes, denoted by $n(E)$, is $3$.

Step 3: Apply the Probability Formula
The probability of an event $P(E)$ is defined as the ratio of the number of favorable outcomes to the total number of possible outcomes:
$P(E) = \frac{n(E)}{n(S)}$
[Using the classical definition of probability]

Step 4: Calculation
Substitute the values identified in Step 1 and Step 2 into the formula:
$P(E) = \frac{3}{6}$
Simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 3:
$P(E) = \frac{3 \div 3}{6 \div 3} = \frac{1}{2}$

Final Answer: The probability of getting a number lying between 2 and 6 is $\frac{1}{2}$.


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