Question

question 6 points 3 what is the probability of throwing an odd number or a non-prime number in a single throw of a fair 6-sided die? options: \\(\frac{2}{3}\\), \\(\frac{5}{6}\\), 1, \\(\frac{1}{2}\\)

Explanation:

Step1: Identify possible outcomes

A fair 6 - sided die has outcomes: \(1, 2, 3, 4, 5, 6\) (total \(n = 6\) outcomes).

Step2: Define events

Let \(A\) be the event of getting an odd number. \(A=\{1, 3, 5\}\), so \(n(A)=3\).
Let \(B\) be the event of getting a non - prime number. Prime numbers between 1 - 6 are \(2, 3, 5\). So non - prime numbers: \(B = \{1, 4, 6\}\), \(n(B)=3\).

Step3: Find \(A\cap B\) (intersection)

\(A\cap B=\{1\}\) (numbers that are odd and non - prime), so \(n(A\cap B) = 1\).

Step4: Use the formula for \(P(A\cup B)\)

The formula for the probability of the union of two events is \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)
\(P(A)=\frac{n(A)}{n}=\frac{3}{6}\), \(P(B)=\frac{n(B)}{n}=\frac{3}{6}\), \(P(A\cap B)=\frac{n(A\cap B)}{n}=\frac{1}{6}\)
\(P(A\cup B)=\frac{3}{6}+\frac{3}{6}-\frac{1}{6}=\frac{3 + 3-1}{6}=\frac{5}{6}\)

Answer:

\(\frac{5}{6}\) (corresponding to the option with \(\frac{5}{6}\))