Question

andrea rolls a number cube twice. she determines ( p(\text{even, then odd}) = \frac{1}{2} ). which statements are accurate? check all that apply.
(square) andrea’s solution is incorrect.
(square) the total number of possible outcomes is 12.
(square) the probability of each simple event is (\frac{1}{2}).
(square) there are three possible outcomes on each number cube.
(square) the probability of the compound event is less than the probability of either event occurring alone.
(square) ( p(\text{even, then odd}) = p(\text{odd, then even}) )

Explanation:

Step1: Analyze total outcomes

A number cube has 6 faces. Rolling it twice, total outcomes = \(6\times6 = 36\). So "The total number of possible outcomes is 12" is wrong.

Step2: Probability of even/odd

On a number cube, even numbers (2,4,6) and odd numbers (1,3,5) are 3 each. So \(P(\text{even})=\frac{3}{6}=\frac{1}{2}\), \(P(\text{odd})=\frac{3}{6}=\frac{1}{2}\). So "The probability of each simple event is \(\frac{1}{2}\)" is correct.

Step3: Andrea's solution

\(P(\text{even, then odd}) = P(\text{even})\times P(\text{odd})=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}
eq\frac{1}{2}\). So Andrea's solution is incorrect (correct statement).

Step4: Outcomes per cube

A number cube has 6 outcomes, not 3. So "There are three possible outcomes on each number cube" is wrong.

Step5: Compound event probability

\(P(\text{compound})=\frac{1}{4}\), \(P(\text{simple})=\frac{1}{2}\). \(\frac{1}{4}<\frac{1}{2}\), so "The probability of the compound event is less than the probability of either event occurring alone" is correct.

Step6: \(P(\text{even, then odd})\) vs \(P(\text{odd, then even})\)

\(P(\text{even, then odd})=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}\), \(P(\text{odd, then even})=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}\). So they are equal (correct statement).

Answer:

• Andrea’s solution is incorrect.
• The probability of each simple event is \(\frac{1}{2}\).
• The probability of the compound event is less than the probability of either event occurring alone.
• \(P(\text{even, then odd}) = P(\text{odd, then even})\)