Question

Jon, an aspiring meteorologist, spent the past few weekends studying the clouds. He took detailed notes on the types of clouds observed and the time of day they were observed.
Stratocumulus Stratus
Morning 6 4
Afternoon 4 3
Evening 2 3
What is the probability that a randomly selected cloud was marked as stratus or was not observed in the afternoon?
Simplify any fractions.

Answer 1

To find the probability that a randomly selected cloud was marked as stratus or was not observed in the afternoon, we first need to calculate the total number of clouds observed and then apply the principles of probability.

### Total Observations

From Jon's notes, we can summarize the observations as follows:

- **Morning**:
- Stratocumulus: 6
- Stratus: 4
- **Afternoon**:
- Stratocumulus: 4
- Stratus: 3
- **Evening**:
- Stratocumulus: 2
- Stratus: 3

Now, let's calculate the total observations.

1. **Morning** = 6 Stratocumulus + 4 Stratus = 10
2. **Afternoon** = 4 Stratocumulus + 3 Stratus = 7
3. **Evening** = 2 Stratocumulus + 3 Stratus = 5

Now, sum all observations:
\[
\text{Total Clouds} = 10 + 7 + 5 = 22
\]

### Identifying the Events

Next, we need to define two events based on the question:

- **Event A**: The cloud was marked as stratus.
- **Event B**: The cloud was not observed in the afternoon.

#### Probability of Event A (Cloud marked as stratus)

The total number of stratus clouds observed is:
- Morning: 4
- Afternoon: 3
- Evening: 3

So the total number of stratus clouds is:
\[
4 + 3 + 3 = 10
\]

Thus, the probability of Event A is:
\[
P(A) = \frac{\text{Number of stratus clouds}}{\text{Total clouds}} = \frac{10}{22} = \frac{5}{11}
\]

#### Probability of Event B (Cloud not observed in the afternoon)

The number of clouds observed in the afternoon is 7 (4 Stratocumulus + 3 Stratus), implying that the clouds not observed in the afternoon are those observed in the morning and evening:
\[
10 \text{ (Morning)} + 5 \text{ (Evening)} = 15
\]

Thus, the probability of Event B is:
\[
P(B) = \frac{\text{Number of clouds not observed in the afternoon}}{\text{Total clouds}} = \frac{15}{22}
\]

#### Probability of Event A and B (Cloud is stratus and not in the afternoon)

The stratus clouds that were observed in the afternoon are counted as well (3). Therefore, the stratus clouds not observed in the afternoon are:
- Morning: 4
- Evening: 3

So, total stratus clouds not observed in the afternoon:
\[
4 + 3 = 7
\]

Thus, the probability of Event A and B is:
\[
P(A \cap B) = \frac{7}{22}
\]

### Using the Inclusion-Exclusion Principle

Now, we find the probability that a randomly selected cloud was marked as stratus or was not observed in the afternoon using the inclusion-exclusion principle:
\[
P(A \cup B) = P(A) + P(B) - P(A \cap B)
\]
Substituting in the probabilities we calculated:
\[
P(A \cup B) = \frac{10}{22} + \frac{15}{22} - \frac{7}{22} = \frac{10 + 15 - 7}{22} = \frac{18}{22} = \frac{9}{11}
\]

### Final Answer

Thus, the probability that a randomly selected cloud was marked as stratus or was not observed in the afternoon is \(\frac{9}{11}\).