Concept explainers
a.
1. Find the
2. Find the probability that the home is in township 1or has a pool.
3. Find the probability that the home has a pool given that it is in township 3.
4. Find the probability that the home has a pool and is in township 3.
a.
Answer to Problem 92DA
- 1. The probability that the home has a pool is 0.64.
- 2. The probability that the home is in township 1or has a pool is 0.72.
- 3. The probability that the home has a pool given that it is in township 3 is 0.72.
- 4. 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 does 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 has a pool:
The probability that the home has a pool is obtained as follows:
Thus, the probability that the home has a pool is 0.64.
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 92DA
- 1. The probability that the home has a garage attached is 0.74.
- 2. The probability that the home does not have a garage attached given that it is township 5 is 0.1875.
- 3. The probability that the home has a garage attached and is in township 3 is 0.18.
- 4. The probability that the home does not have a garage attached or is in township 2is 0.39.
Explanation of Solution
Calculation:
The below table shows the number of homes that have a garage versus the number of homes that does not have a garage in each of the five townships.
Township | ||||||
garage | 1 | 2 | 3 | 4 | 5 | Total |
Yes | 9 | 14 | 19 | 23 | 13 | 78 |
No | 6 | 6 | 6 | 6 | 3 | 27 |
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 0.74.
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 0.1875.
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 0.18.
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 0.39.
Want to see more full solutions like this?
Chapter 5 Solutions
Gen Combo Ll Statistical Techniques In Business And Economics; Connect Ac
- T1.4: Let ẞ(G) be the minimum size of a vertex cover, a(G) be the maximum size of an independent set and m(G) = |E(G)|. (i) Prove that if G is triangle free (no induced K3) then m(G) ≤ a(G)B(G). Hints - The neighborhood of a vertex in a triangle free graph must be independent; all edges have at least one end in a vertex cover. (ii) Show that all graphs of order n ≥ 3 and size m> [n2/4] contain a triangle. Hints - you may need to use either elementary calculus or the arithmetic-geometric mean inequality.arrow_forwardWe consider the one-period model studied in class as an example. Namely, we assumethat the current stock price is S0 = 10. At time T, the stock has either moved up toSt = 12 (with probability p = 0.6) or down towards St = 8 (with probability 1−p = 0.4).We consider a call option on this stock with maturity T and strike price K = 10. Theinterest rate on the money market is zero.As in class, we assume that you, as a customer, are willing to buy the call option on100 shares of stock for $120. The investor, who sold you the option, can adopt one of thefollowing strategies: Strategy 1: (seen in class) Buy 50 shares of stock and borrow $380. Strategy 2: Buy 55 shares of stock and borrow $430. Strategy 3: Buy 60 shares of stock and borrow $480. Strategy 4: Buy 40 shares of stock and borrow $280.(a) For each of strategies 2-4, describe the value of the investor’s portfolio at time 0,and at time T for each possible movement of the stock.(b) For each of strategies 2-4, does the investor have…arrow_forwardNegate the following compound statement using De Morgans's laws.arrow_forward
- Negate the following compound statement using De Morgans's laws.arrow_forwardQuestion 6: Negate the following compound statements, using De Morgan's laws. A) If Alberta was under water entirely then there should be no fossil of mammals.arrow_forwardNegate the following compound statement using De Morgans's laws.arrow_forward
- Characterize (with proof) all connected graphs that contain no even cycles in terms oftheir blocks.arrow_forwardLet G be a connected graph that does not have P4 or C3 as an induced subgraph (i.e.,G is P4, C3 free). Prove that G is a complete bipartite grapharrow_forwardProve sufficiency of the condition for a graph to be bipartite that is, prove that if G hasno odd cycles then G is bipartite as follows:Assume that the statement is false and that G is an edge minimal counterexample. That is, Gsatisfies the conditions and is not bipartite but G − e is bipartite for any edge e. (Note thatthis is essentially induction, just using different terminology.) What does minimality say aboutconnectivity of G? Can G − e be disconnected? Explain why if there is an edge between twovertices in the same part of a bipartition of G − e then there is an odd cyclearrow_forward
- Holt Mcdougal Larson Pre-algebra: Student Edition...AlgebraISBN:9780547587776Author:HOLT MCDOUGALPublisher:HOLT MCDOUGALCollege Algebra (MindTap Course List)AlgebraISBN:9781305652231Author:R. David Gustafson, Jeff HughesPublisher:Cengage Learning
- Algebra & Trigonometry with Analytic GeometryAlgebraISBN:9781133382119Author:SwokowskiPublisher:CengageGlencoe Algebra 1, Student Edition, 9780079039897...AlgebraISBN:9780079039897Author:CarterPublisher:McGraw Hill