you roll a 6 - sided die. what is p(odd or less than 2)? write your answer as a percentage.

Explanation:

Step1: Define the sample space.

A 6-sided die has possible outcomes $S = {1, 2, 3, 4, 5, 6}$. The total number of outcomes is $n(S) = 6$.

Step2: Identify outcomes for "odd".

Let A be the event of rolling an odd number. The odd outcomes are $A = {1, 3, 5}$. So, $n(A) = 3$.

Step3: Identify outcomes for "less than 2".

Let B be the event of rolling a number less than 2. The outcome is $B = {1}$. So, $n(B) = 1$.

Step4: Identify outcomes for "odd OR less than 2".

This is the union of events A and B, denoted as $A \cup B$. $A \cup B = {1, 3, 5} \cup {1} = {1, 3, 5}$. The number of favorable outcomes is $n(A \cup B) = 3$.

Step5: Calculate the probability.

The probability P(odd or less than 2) is $\frac{n(A \cup B)}{n(S)}$. $$P(\text{odd or less than 2}) = \frac{3}{6} = \frac{1}{2}$$

Step6: Convert probability to percentage.

To express the probability as a percentage, multiply by 100. $$\text{Percentage} = \frac{1}{2} \times 100% = 50%$$

Answer:

50%