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 6.
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 5 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.
{1H, 1T, 2H, 2T, 3H, 3T, 4H, 4T, 5H, 5T, 6H, 6T}

Step2: Calculate probability of getting a head

There are 12 total outcomes in the sample - space. The number of outcomes with a head is 6. Using the probability formula $P(A)=\frac{n(A)}{n(S)}$, where $n(A)$ is the number of favorable outcomes and $n(S)$ is the number of total outcomes.
$P(\text{head})=\frac{6}{12}=\frac{1}{2}$

Step3: Calculate probability of getting a 6

The outcomes with a 6 are 6H and 6T. So $n(A) = 2$ and $n(S)=12$.
$P(6)=\frac{2}{12}=\frac{1}{6}$

Step4: Calculate probability of getting a 3 or a head

The number of outcomes with a 3 are 3H and 3T (2 outcomes). The number of outcomes with a head is 6. But 3H is counted twice. Using the inclusion - exclusion principle $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. Here, $A$ is the event of getting a 3 and $B$ is the event of getting a head. $P(A)=\frac{2}{12}$, $P(B)=\frac{6}{12}$, and $P(A\cap B)=\frac{1}{12}$.
$P(3\text{ or head})=\frac{2 + 6-1}{12}=\frac{7}{12}$

Step5: Calculate probability of getting a 5 and a tail

The outcome is 5T, so $n(A) = 1$ and $n(S)=12$.
$P(5\text{ and tail})=\frac{1}{12}$

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) $\frac{1}{12}$