A dice is thrown once. Then the probability of "odd number" will be:

Question

Accepted Answer

The correct option is 1:

To solve this problem, we need to understand the basic concept of probability. The probability of an event is defined as the ratio of the number of favorable outcomes to the total number of possible outcomes.
The formula for probability is given by:
$P(E) = \frac{	ext{Number of favorable outcomes}}{	ext{Total number of possible outcomes}}$

Step 1: Identify the total number of possible outcomes when a dice is thrown once.
A standard six-sided die has the following possible outcomes: {1, 2, 3, 4, 5, 6}.
So, the total number of possible outcomes, $n(S)$, is 6.

Step 2: Identify the favorable outcomes, which are the "odd numbers".
From the possible outcomes {1, 2, 3, 4, 5, 6}, the odd numbers are {1, 3, 5}.
So, the number of favorable outcomes (getting an odd number), $n(E)$, is 3.

Step 3: Calculate the probability of getting an odd number.
Using the probability formula:
$P(	ext{odd number}) = \frac{	ext{Number of odd numbers}}{	ext{Total number of outcomes}}$
$P(	ext{odd number}) = \frac{3}{6}$
$P(	ext{odd number}) = \frac{1}{2}$

Thus, the probability of throwing an odd number when a dice is thrown once is $\frac{1}{2}$. This matches Option 1.

The incorrect option is 2 (0): This would imply that it's impossible to roll an odd number, which is incorrect as 1, 3, and 5 are odd numbers on a die.
The incorrect option is 3 ($\frac{1}{3}$): This would be the probability if there were 2 favorable outcomes out of 6 (e.g., rolling a 3 or a 6), which is not the case for odd numbers.
The incorrect option is 4 ($\frac{2}{3}$): This would be the probability if there were 4 favorable outcomes out of 6 (e.g., rolling numbers greater than 2), which is not the case for odd numbers.