Concept explainers
a.
1. Find the
2. Find the probability that the home has a pool given that it is in township 3.
3. Find the probability that the home has a pool and is in township 3.
a.
Answer to Problem 92DE
- 1. The probability that the home is in township 1or has a pool is 0.72.
- 2. The probability that the home has a pool given that it is in township 3 is 0.72.
- 3. The probability that the home has a pool and is in township 3 is 0.17.
Explanation of Solution
Calculation:
The below table shows the number of homes that have a pool, versus the number of homes that do not have a pool, in each of the five townships.
Township | ||||||
pool | 1 | 2 | 3 | 4 | 5 | Total |
Yes | 6 | 12 | 18 | 18 | 13 | 67 |
No | 9 | 8 | 7 | 11 | 3 | 38 |
Total | 15 | 20 | 25 | 29 | 16 | 105 |
Home is in township 1 or has a pool:
The probability that the home is in township 1 or has a pool is obtained as follows:
Thus, the probability that the home is in township 1 or has a pool is 0.72.
Home has a pool given that it is in township 3:
The probability that the home has a pool given that it is in township 3 is obtained as follows:
Thus, the probability that the home has a pool given that it is in township 3 is 0.72.
Home has a pool and is in township 3:
The probability that the home has a pool and is in township 3 is obtained as follows:
Thus, the probability that the home has a pool and is in township 3 is 0.17.
b.
1. Find the probability that the home has a garage attached.
2. Find the probability that the home does not have a garage attached given that it is township 5.
3. Find the probability that the home has a garage attached and is in township 3.
4. Find the probability that the home does not have a garage attached or is in township 2.
b.
Answer to Problem 92DE
- 1. The probability that the home has a garage attached is 0.68.
- 2. The probability that the home does not have a garage attached given that it is township 5 is 0.25.
- 3. The probability that the home has a garage attached and is in township 3 is 0.14.
- 4. The probability that the home does not have a garage attached or is in township 2is 0.47.
Explanation of Solution
Calculation:
The below table shows the number of homes that have a garage versus the number of homes that do not have a garage in each of the five townships.
Township | ||||||
garage | 1 | 2 | 3 | 4 | 5 | Total |
Yes | 9 | 15 | 15 | 20 | 12 | 71 |
No | 6 | 5 | 10 | 9 | 4 | 34 |
Total | 15 | 20 | 25 | 29 | 16 | 105 |
Home has a garage attached:
The probability that the home has a garage attached is obtained as follows:
Thus, the probability that the home has a garage attached is 0.68.
Home does not have a garage attached given that it is in township 5:
The probability that the home does not have a garage attached given that it is in township 5 is obtained as follows:
Thus, the probability that the home does not have a garage attached given that it is in township 5 is 0.25.
Home has a garage attached and is in township 3:
The probability that the home has a garage attached and is in township 3 is obtained as follows:
Thus, the probability that the home has a garage attached and is in township 3 is 0.14.
Home does not have a garage attached or in township 2:
The probability that the home does not have a garage attached or in township 2 is obtained as follows:
Thus, the probability that the home does not have a garage attached or in township 2 is 0.47.
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Chapter 5 Solutions
Loose Leaf for Statistical Techniques in Business and Economics (Mcgraw-hill/Irwin Series in Operations and Decision Sciences)
- 6. Show that 1{AU B} = max{1{A}, I{B}} = I{A} + I{B} - I{A} I{B}; I{AB} = min{I{A}, I{B}} = I{A} I{B}; I{A A B} = I{A} + I{B}-21{A} I {B} = (I{A} - I{B})².arrow_forwardTheorem 3.5 Suppose that P and Q are probability measures defined on the same probability space (2, F), and that F is generated by a л-system A. If P(A) = Q(A) for all A = A, then P = Q, i.e., P(A) = Q(A) for all A = F.arrow_forward6. Show that, for any random variable, X, and a > 0, Lo P(x -00 P(x < xarrow_forward5. Suppose that X is an integer valued random variable, and let mЄ N. Show that 8 11118 P(narrow_forward食食假 6. Show that I(AUB) = max{1{A}, I{B}} = I{A} + I{B} - I{A} I{B}; I(AB)= min{I{A}, I{B}} = I{A} I{B}; I{A A B} = I{A} + I{B}-21{A} I{B} = (I{A} - I{B})². -arrow_forward11. Suppose that the events (An, n ≥ 1) are independent. Show that the inclusion- exclusion formula reduces to P(UAL)-1-(1-P(Ak)). k=1 k=1arrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_iosRecommended textbooks for you
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