Question

an experiment is rolling a fair die and then flipping a coin.
a) state the sample space. list outcomes separated by commas. use notation like 1h to mean you rolled a 1 on the die and flipped heads on the coin.
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b) find the probability of getting a head.
leave your answer as a fraction.

c) find the probability of getting a 5.
leave your answer as a fraction.

d) find the probability of getting a 3 or a head.
leave your answer as a fraction.

e) find the probability of getting a 4 and a tail.
leave your answer as a fraction.

question help: video 1 video 2 message instructor

Explanation:

Step1: Determine sample - space

A die has 6 faces and a coin has 2 sides. The sample - space is the set of all possible combined outcomes. So, the sample space $S=\{1H,1T,2H,2T,3H,3T,4H,4T,5H,5T,6H,6T\}$.

Step2: Calculate probability of getting a head

The total number of outcomes in the sample space $n(S)=12$. The number of outcomes with a head $n(H) = 6$. Using the probability formula $P(A)=\frac{n(A)}{n(S)}$, the probability of getting a head $P(H)=\frac{6}{12}=\frac{1}{2}$.

Step3: Calculate probability of getting a 5

The number of outcomes with a 5 (either 5H or 5T) is $n(5)=2$. So, the probability of getting a 5, $P(5)=\frac{2}{12}=\frac{1}{6}$.

Step4: Calculate probability of getting a 3 or a head

Let $A$ be the event of getting a 3 and $B$ be the event of getting a head. $n(A) = 2$ (3H and 3T), $n(B)=6$, and $n(A\cap B)=1$ (3H). Using the formula $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, $P(A)=\frac{2}{12}$, $P(B)=\frac{6}{12}$, $P(A\cap B)=\frac{1}{12}$. Then $P(A\cup B)=\frac{2 + 6-1}{12}=\frac{7}{12}$.

Step5: Calculate probability of getting a 4 and a tail

There is only 1 outcome that is 4T. So, the probability $P(4T)=\frac{1}{12}$.

Answer:

a) $1H,1T,2H,2T,3H,3T,4H,4T,5H,5T,6H,6T$
b) $\frac{1}{2}$
c) $\frac{1}{6}$
d) $\frac{7}{12}$
e) $\frac{1}{12}$