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.
andreas solution is incorrect.
the total number of possible outcomes is 12.
the probability of each simple event is ( \frac{1}{2} ).
there are three possible outcomes on each number cube.
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}) )

Explanation:

Step1: Calculate total number of outcomes

When rolling a number cube twice, by the fundamental counting principle, the total number of possible outcomes is \(n = 6\times6=36
eq12\).

Step2: Calculate probability of simple event

For a single - roll of a number cube, \(P(\text{even})=\frac{3}{6}=\frac{1}{2}\) and \(P(\text{odd})=\frac{3}{6}=\frac{1}{2}\).

Step3: Calculate \(P(\text{even, then odd})\)

Since the two rolls are independent events, \(P(A\cap B)=P(A)\times P(B)\). Here \(A\) is getting an even number on the first roll and \(B\) is getting an odd number on the second roll. So \(P(\text{even, then odd})=\frac{3}{6}\times\frac{3}{6}=\frac{9}{36}=\frac{1}{4}
eq\frac{1}{2}\), so Andrea's solution is incorrect.

Step4: Analyze \(P(\text{odd, then even})\)

\(P(\text{odd, then even})=\frac{3}{6}\times\frac{3}{6}=\frac{9}{36}=\frac{1}{4}\), so \(P(\text{even, then odd}) = P(\text{odd, then even})\)

Step5: Compare probabilities

\(P(\text{even})=\frac{1}{2}\), \(P(\text{odd})=\frac{1}{2}\), and \(P(\text{even, then odd})=\frac{1}{4}\). So the probability of the compound event is less than the probability of either event occurring alone.

Step6: Check number of outcomes on a number cube

A standard number cube has \(6\) possible outcomes (\(1,2,3,4,5,6\)), not \(3\).

Answer:

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