Question

probability and tree diagrams
use a probability tree to help answer the following questions.
a regular die has 6 sides and a coin has 2 sides. you roll the die once then flip the coin once.
what is the probability of:

1. rolling a 5 and flipping a heads?
2. rolling a 5 or flipping a heads?
3. rolling an even number and flipping a tails?
4. rolling an even number or flipping a tails?

a bag has 2 marbles labeled 4 and 7 and a spinner has 3 options. you randomly pull one marble then spin the spinner once.
what is the probability of:

1. drawing a number less than 8 or spinning a 4?
2. drawing a number less than 8 and spinning a 4?
3. drawing an 8 or spinning a 3?
4. drawing an 8 and spinning a 3?

Explanation:

Step1: Recall probability formulas

For independent - events \(A\) and \(B\), \(P(A\cap B)=P(A)\times P(B)\) and \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\).

Step2: Calculate probabilities for die - coin case

1. The probability of rolling a 5 on a 6 - sided die is \(P(\text{roll }5)=\frac{1}{6}\), and the probability of flipping a head on a coin is \(P(\text{head})=\frac{1}{2}\).

• For \(P(\text{roll }5\cap\text{head})\), since they are independent events, \(P(\text{roll }5\cap\text{head})=\frac{1}{6}\times\frac{1}{2}=\frac{1}{12}\).

1. \(P(\text{roll }5)=\frac{1}{6}\), \(P(\text{head})=\frac{1}{2}\), and \(P(\text{roll }5\cap\text{head})=\frac{1}{12}\).

• Using the formula \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\), \(P(\text{roll }5\cup\text{head})=\frac{1}{6}+\frac{1}{2}-\frac{1}{12}=\frac{2 + 6-1}{12}=\frac{7}{12}\).

1. The probability of rolling an even number on a 6 - sided die (\(2,4,6\)) is \(P(\text{even})=\frac{3}{6}=\frac{1}{2}\), and the probability of flipping a tail on a coin is \(P(\text{tail})=\frac{1}{2}\).

• For \(P(\text{even}\cap\text{tail})\), since they are independent events, \(P(\text{even}\cap\text{tail})=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}\).

1. \(P(\text{even})=\frac{1}{2}\), \(P(\text{tail})=\frac{1}{2}\), and \(P(\text{even}\cap\text{tail})=\frac{1}{4}\).

• Using the formula \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\), \(P(\text{even}\cup\text{tail})=\frac{1}{2}+\frac{1}{2}-\frac{1}{4}=\frac{4 - 1}{4}=\frac{3}{4}\).

Step3: Calculate probabilities for marble - spinner case

The probability of drawing a number less than 8 from the bag (both 4 and 7 are less than 8) is \(P(\text{less than }8)=1\), and the probability of spinning a 4 on a 3 - option spinner is \(P(\text{spin }4)=0\).

1. \(P(\text{less than }8\cup\text{spin }4)=P(\text{less than }8)+P(\text{spin }4)-P(\text{less than }8\cap\text{spin }4)=1 + 0-0 = 1\).
2. \(P(\text{less than }8\cap\text{spin }4)=P(\text{less than }8)\times P(\text{spin }4)=1\times0 = 0\).

The probability of drawing an 8 from the bag is \(P(\text{draw }8)=0\), and the probability of spinning a 3 on a 3 - option spinner is \(P(\text{spin }3)=\frac{1}{3}\).

1. \(P(\text{draw }8\cup\text{spin }3)=P(\text{draw }8)+P(\text{spin }3)-P(\text{draw }8\cap\text{spin }3)=0+\frac{1}{3}-0=\frac{1}{3}\).
2. \(P(\text{draw }8\cap\text{spin }3)=P(\text{draw }8)\times P(\text{spin }3)=0\times\frac{1}{3}=0\).

Answer:

1. \(\frac{1}{12}\)
2. \(\frac{7}{12}\)
3. \(\frac{1}{4}\)
4. \(\frac{3}{4}\)
5. \(1\)
6. \(0\)
7. \(\frac{1}{3}\)
8. \(0\)