Question

4 suppose you roll a dice and then flip a coin. a. create a probability area model or tree diagram to represent this situation. b. what is the probability you roll a 2 and then flip a heads? c. what is the probability you roll a 3 or a 4? d. what is the probability you roll a 6 and then flip a tails? e. what is the probability you roll an 8? f. what is the probability you flip a heads or a tails? g. what is the probability you roll a 3 or flip a tails?

Explanation:

Step1: Identify probabilities of single - events

The probability of rolling a specific number on a fair six - sided die is $P(\text{any number on die})=\frac{1}{6}$, and the probability of getting heads or tails on a fair coin flip is $P(\text{heads}) = P(\text{tails})=\frac{1}{2}$.

Step2: For part b

The events of rolling a die and flipping a coin are independent. For independent events $A$ and $B$, $P(A\cap B)=P(A)\times P(B)$. Let $A$ be the event of rolling a 2 on the die ($P(A)=\frac{1}{6}$) and $B$ be the event of flipping a heads ($P(B)=\frac{1}{2}$). Then $P(A\cap B)=\frac{1}{6}\times\frac{1}{2}=\frac{1}{12}$.

Step3: For part c

The probability of rolling a 3 or a 4 on a die. Using the addition rule for mutually - exclusive events $P(A\cup B)=P(A)+P(B)$. Let $A$ be the event of rolling a 3 ($P(A)=\frac{1}{6}$) and $B$ be the event of rolling a 4 ($P(B)=\frac{1}{6}$). Then $P(A\cup B)=\frac{1}{6}+\frac{1}{6}=\frac{1}{3}$.

Step4: For part d

Let $A$ be the event of rolling a 6 on the die ($P(A)=\frac{1}{6}$) and $B$ be the event of flipping a tails ($P(B)=\frac{1}{2}$). Since they are independent, $P(A\cap B)=\frac{1}{6}\times\frac{1}{2}=\frac{1}{12}$.

Step5: For part e

A standard six - sided die has values from 1 to 6. So the probability of rolling an 8 is $P = 0$.

Step6: For part f

The probability of flipping a heads or a tails on a coin. Since these are the only two possible outcomes of a coin flip, $P(\text{heads}\cup\text{tails})=\frac{1}{2}+\frac{1}{2}=1$.

Step7: For part g

Let $A$ be the event of rolling a 3 on the die ($P(A)=\frac{1}{6}$) and $B$ be the event of flipping a tails ($P(B)=\frac{1}{2}$). Using the addition rule $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. Since $A$ and $B$ are independent, $P(A\cap B)=\frac{1}{6}\times\frac{1}{2}=\frac{1}{12}$. Then $P(A\cup B)=\frac{1}{6}+\frac{1}{2}-\frac{1}{12}=\frac{2 + 6-1}{12}=\frac{7}{12}$.

Answer:

b. $\frac{1}{12}$
c. $\frac{1}{3}$
d. $\frac{1}{12}$
e. $0$
f. $1$
g. $\frac{7}{12}$