Question

two spinners are each spun one time. make a probability tree or sample space to show all the combinations of the two spinners above. what is the probability of: 1) spinning the same number twice? 2) spinning two odd numbers? 3) spinning a 1 or a 2? of two marbles numbered 4 and 5 in a bag, one is chosen blindly. also, one die with six sides, numbered 1 through 6, is rolled once. make a probability tree or sample space to show all the combinations of the marbles and die. what is the probability of: 4) pulling a 4 or rolling a 6? 5) pulling an even number and rolling an odd number? 6) pulling a 5 or rolling a prime number? 7) pulling a 3 or rolling a 4?

Explanation:

Step1: Find total outcomes for spinners

The first spinner has 3 outcomes and the second has 4 outcomes. By the fundamental - counting principle, the total number of outcomes when spinning both spinners is \(3\times4 = 12\).

Step2: Probability of spinning the same number twice

The pairs where the numbers are the same are \((1,1)\), \((2,2)\), \((3,3)\). There are 2 such pairs (since the first spinner doesn't have 4). So the probability is \(\frac{2}{12}=\frac{1}{6}\).

Step3: Probability of spinning two odd numbers

Odd - numbered outcomes for the first spinner are 1 and 3, and for the second spinner are 1 and 3. The pairs of odd numbers are \((1,1)\), \((1,3)\), \((3,1)\), \((3,3)\). So there are 4 pairs, and the probability is \(\frac{4}{12}=\frac{1}{3}\).

Step4: Probability of spinning a 1 or a 2

The number of pairs with 1 or 2:
For the first spinner, when it shows 1: \((1,1)\), \((1,2)\), \((1,3)\), \((1,4)\); when it shows 2: \((2,1)\), \((2,2)\), \((2,3)\), \((2,4)\). There are 8 such pairs. So the probability is \(\frac{8}{12}=\frac{2}{3}\).

Step5: Find total outcomes for marbles and die

The marble selection has 2 outcomes and the die - roll has 6 outcomes. So the total number of combined outcomes is \(2\times6=12\).

Step6: Probability of pulling a 4 or rolling a 6

The number of outcomes of pulling a 4 is 1, and the number of outcomes of rolling a 6 is 1. The number of outcomes of pulling a 4 and rolling a 6 is 1. Using the formula \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\), \(P(A)=\frac{1}{2}\), \(P(B)=\frac{1}{6}\), \(P(A\cap B)=\frac{1}{12}\). So \(P(A\cup B)=\frac{1}{2}+\frac{1}{6}-\frac{1}{12}=\frac{6 + 2-1}{12}=\frac{7}{12}\).

Step7: Probability of pulling an even number and rolling an odd number

The even - numbered marble is 4. The odd numbers on the die are 1, 3, 5. So there are 3 such pairs, and the probability is \(\frac{3}{12}=\frac{1}{4}\).

Step8: Probability of pulling a 5 or rolling a prime number

The number of outcomes of pulling a 5 is 1, the prime numbers on the die are 2, 3, 5. There are 3 prime numbers on the die. The number of outcomes of pulling a 5 and rolling a prime number is 1 (when rolling 5). Using the formula \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\), \(P(A)=\frac{1}{2}\), \(P(B)=\frac{3}{6}=\frac{1}{2}\), \(P(A\cap B)=\frac{1}{12}\). So \(P(A\cup B)=\frac{1}{2}+\frac{1}{2}-\frac{1}{12}=\frac{11}{12}\).

Step9: Probability of pulling a 3 or rolling a 4

The number of outcomes of pulling a 3 is 0 (since there is no 3 in the marbles), and the number of outcomes of rolling a 4 is 1. So the probability is \(\frac{1}{12}\).

Answer:

1. \(\frac{1}{6}\)
2. \(\frac{1}{3}\)
3. \(\frac{2}{3}\)
4. \(\frac{7}{12}\)
5. \(\frac{1}{4}\)
6. \(\frac{11}{12}\)
7. \(\frac{1}{12}\)