Question

andrea rolls a number cube twice. she determines p(even, then odd) = 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 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(even, then odd) = p(odd, then even)

Explanation:

Step1: Calculate total outcomes

When rolling a number - cube twice, by the fundamental counting principle, the total number of possible outcomes is \(6\times6 = 36\), not 12.

Step2: Calculate probability of simple events

The probability of getting an even number (\(2\), \(4\), \(6\)) on a single roll of a number - cube is \(P(\text{even})=\frac{3}{6}=\frac{1}{2}\), and the probability of getting an odd number (\(1\), \(3\), \(5\)) on a single roll is \(P(\text{odd})=\frac{3}{6}=\frac{1}{2}\).

Step3: Calculate compound - event probability

Since the two rolls are independent events, \(P(\text{even, then odd})=P(\text{even})\times P(\text{odd})=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}
eq\frac{1}{2}\), so Andrea's solution is incorrect. Also, \(P(\text{odd, then even}) = P(\text{odd})\times P(\text{even})=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}\), so \(P(\text{even, then odd})=P(\text{odd, then even})\). The probability of each simple event (even or odd on a single roll) is \(\frac{1}{2}\), and there are 6 possible outcomes on each number - cube. The probability of the compound event \(P(\text{even, then odd})=\frac{1}{4}\) which is less than the probability of either event occurring alone (\(\frac{1}{2}\)).

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.
\(P(\text{even, then odd}) = P(\text{odd, then even})\)