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andrea rolls a number cube twice. she determines ( p(\text{even, then o…

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)

  • andrea’s 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: Analyze total outcomes

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

Step2: Probability of simple event

Probability of even (or odd) on a cube: \(\frac{3}{6}=\frac{1}{2}\), so this is correct.

Step3: Number of outcomes per cube

A cube has 6 outcomes (1 - 6), so "three outcomes" is wrong.

Step4: Compound event probability

\(P(\text{even then odd})=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}\), but Andrea said \(\frac{1}{2}\), so her solution is incorrect. Also, \(P(\text{even then odd}) = P(\text{odd then even})=\frac{1}{4}\), so that statement is correct. And \(\frac{1}{4}<\frac{1}{2}\), so "compound event prob less than single" is correct.

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