Question
To find the probability that the school is closed tomorrow using the tree diagram, follow these steps:
Identify Paths to Closure (C): Look for all paths in the tree diagram that lead to the school being closed (C).
Calculate Path Probabilities: Multiply the probabilities along each path that leads to closure. For example, if one path is H (heavy snowfall) leading to C (closed), and the probabilities are 0.7 for H and 0.9 for C given H, then the probability for this path is
0.7
×
0.9
=
0.63
0.7×0.9=0.63.
Sum the Probabilities: Add the probabilities of all paths that result in the school being closed. This gives the total probability of closure.
Identify Paths to Closure (C): Look for all paths in the tree diagram that lead to the school being closed (C).
Calculate Path Probabilities: Multiply the probabilities along each path that leads to closure. For example, if one path is H (heavy snowfall) leading to C (closed), and the probabilities are 0.7 for H and 0.9 for C given H, then the probability for this path is
0.7
×
0.9
=
0.63
0.7×0.9=0.63.
Sum the Probabilities: Add the probabilities of all paths that result in the school being closed. This gives the total probability of closure.
Answers
Answered by
GPT-4o mini
To find the probability that the school is closed tomorrow using a tree diagram, follow these steps outlined clearly:
### Step 1: Identify Paths to Closure (C)
Review the tree diagram for all possible outcomes leading to the school being closed (denote that as C). For instance, potential paths might include:
1. Heavy Snowfall (H) leading to Closure (C)
2. Light Snowfall (L) leading to Closure (C)
3. Sunny (S) leading to Closure (C) (if applicable, depending on conditions)
### Step 2: Calculate Path Probabilities
For each identified path, calculate the probability by multiplying the probabilities along that path. For example:
- Path 1: If the path is H leading to C:
- Probability of Heavy Snowfall (H) = 0.7
- Probability of Closure given Heavy Snowfall (C|H) = 0.9
Thus, the probability for this path is:
\[
P(H) \times P(C|H) = 0.7 \times 0.9 = 0.63
\]
- Path 2: If another path is L leading to C:
- Probability of Light Snowfall (L) = 0.2
- Probability of Closure given Light Snowfall (C|L) = 0.5
Thus, the probability for this path is:
\[
P(L) \times P(C|L) = 0.2 \times 0.5 = 0.1
\]
- Path 3: If somehow Sunny (S) can also lead to closure:
- Probability of Sunny (S) = 0.1
- Probability of Closure given Sunny (C|S) = 0.1
Thus, the probability for this path is:
\[
P(S) \times P(C|S) = 0.1 \times 0.1 = 0.01
\]
### Step 3: Sum the Probabilities
Now, add the probabilities of all the paths leading to closure to get the total probability:
\[
Total \, Probability (C) = P(H \, leading \, to \, C) + P(L \, leading \, to \, C) + P(S \, leading \, to \, C)
\]
Using the examples above:
\[
Total \, Probability (C) = 0.63 + 0.1 + 0.01 = 0.74
\]
### Conclusion
The final result, which would be the total probability of closure tomorrow, is:
\[
P(C) = 0.74
\]
Make sure to adjust the probabilities and the paths according to the specific tree diagram and the conditions that lead to the school being closed.
### Step 1: Identify Paths to Closure (C)
Review the tree diagram for all possible outcomes leading to the school being closed (denote that as C). For instance, potential paths might include:
1. Heavy Snowfall (H) leading to Closure (C)
2. Light Snowfall (L) leading to Closure (C)
3. Sunny (S) leading to Closure (C) (if applicable, depending on conditions)
### Step 2: Calculate Path Probabilities
For each identified path, calculate the probability by multiplying the probabilities along that path. For example:
- Path 1: If the path is H leading to C:
- Probability of Heavy Snowfall (H) = 0.7
- Probability of Closure given Heavy Snowfall (C|H) = 0.9
Thus, the probability for this path is:
\[
P(H) \times P(C|H) = 0.7 \times 0.9 = 0.63
\]
- Path 2: If another path is L leading to C:
- Probability of Light Snowfall (L) = 0.2
- Probability of Closure given Light Snowfall (C|L) = 0.5
Thus, the probability for this path is:
\[
P(L) \times P(C|L) = 0.2 \times 0.5 = 0.1
\]
- Path 3: If somehow Sunny (S) can also lead to closure:
- Probability of Sunny (S) = 0.1
- Probability of Closure given Sunny (C|S) = 0.1
Thus, the probability for this path is:
\[
P(S) \times P(C|S) = 0.1 \times 0.1 = 0.01
\]
### Step 3: Sum the Probabilities
Now, add the probabilities of all the paths leading to closure to get the total probability:
\[
Total \, Probability (C) = P(H \, leading \, to \, C) + P(L \, leading \, to \, C) + P(S \, leading \, to \, C)
\]
Using the examples above:
\[
Total \, Probability (C) = 0.63 + 0.1 + 0.01 = 0.74
\]
### Conclusion
The final result, which would be the total probability of closure tomorrow, is:
\[
P(C) = 0.74
\]
Make sure to adjust the probabilities and the paths according to the specific tree diagram and the conditions that lead to the school being closed.