Question

• mastery check: probability •

a spinner has 3 options and a bag has 4 numbered marbles. you spin the spinner once and randomly pull 1 marble.

1. make a probability tree or sample space to show all combinations of the spinner and marbles.
2. what is the probability of spinning the letter ‘a’ or pulling an even number?
3. what is the probability of not spinning the letter ‘b’ and pulling a 5?

a spinner has 8 options and a bag has 9 numbered marbles. you spin the spinner once and randomly pull 1 marble.

1. what is the total number of combinations?
2. what is the probability of spinning the number 8 and pulling an odd number?

challenge:
use a spinner divided into 3 equal parts, a coin, and a die to answer the following question.

1. what is the probability of spinning the number 2, flipping heads, and rolling a 3?

Explanation:

Step1: Calculate total number of outcomes for each part

1. For the first - part with 3 - option spinner and 4 - marbles, total outcomes = 3×4 = 12. For 8 - option spinner and 9 - marbles, total outcomes = 8×9 = 72. For spinner, coin and die, total outcomes = 3×2×6 = 36.

Step2: Solve part 2

Let \(P(A)\) be the probability of spinning 'A' (\(P(A)=\frac{1}{3}\)), and \(P(E)\) be the probability of pulling an even - numbered marble (\(P(E)=\frac{2}{4}=\frac{1}{2}\)). \(P(A\cap E)=\frac{1}{3}\times\frac{2}{4}=\frac{1}{6}\). Using the formula \(P(A\cup E)=P(A)+P(E)-P(A\cap E)\), we have \(P(A\cup E)=\frac{1}{3}+\frac{1}{2}-\frac{1}{6}=\frac{2 + 3-1}{6}=\frac{4}{6}=\frac{2}{3}\).

Step3: Solve part 3

The probability of not spinning 'B' is \(P(\text{not }B)=\frac{2}{3}\), and the probability of pulling a 5 is \(P(5)=\frac{1}{4}\). Since these are independent events, \(P(\text{not }B\cap5)=\frac{2}{3}\times\frac{1}{4}=\frac{1}{6}\).

Step4: Solve part 4

For an 8 - option spinner and 9 - marbles, by the multiplication principle, the total number of combinations is \(8\times9 = 72\).

Step5: Solve part 5

The probability of spinning 8 is \(P(8)=\frac{1}{8}\), and the probability of pulling an odd - numbered marble (\(1,3,5,7,9\)) is \(P(\text{odd})=\frac{5}{9}\). Since these are independent events, \(P(8\cap\text{odd})=\frac{1}{8}\times\frac{5}{9}=\frac{5}{72}\).

Step6: Solve part 6

The probability of spinning 2 is \(P(2)=\frac{1}{3}\), the probability of flipping heads is \(P(H)=\frac{1}{2}\), and the probability of rolling a 3 is \(P(3)=\frac{1}{6}\). Since these are independent events, \(P(2\cap H\cap3)=\frac{1}{3}\times\frac{1}{2}\times\frac{1}{6}=\frac{1}{36}\).

Answer:

1. Sample space has 12 elements (A2, A3, A4, A5, B2, B3, B4, B5, C2, C3, C4, C5) for the first spinner - marble case.
2. \(\frac{2}{3}\)
3. \(\frac{1}{6}\)
4. 72
5. \(\frac{5}{72}\)
6. \(\frac{1}{36}\)