BENG 100 Final

Suppose we have the following information: There is a 60 percent chance that it will rain today. There is a 50 percent chance that it will rain tomorrow. There is a 30 percent chance that it does not rain either day. Find the following probabilities: The probability that it will rain today or tomorrow. The probability that it will rain today and tomorrow. The probability that it will rain today but not tomorrow. The probability that it either will rain today or tomorrow, but not both

An important step in solving problems like this is to correctly convert them to probability language. This is especially useful when the problems become complex. For this problem, let's define A as the event that it will rain today, and B as the event that it will rain tomorrow. Then, let's summarize the available information: P(A)=0.6, P(B)=0.5, P(Ac∩Bc)=0.3 Now that we have summarized the information, we should be able to use them alongside probability rules to find the requested probabilities: The probability that it will rain today or tomorrow: this is P(A∪B). To find this we notice that P(A∪B) =1−P((A∪B)c) by Example 1.10 =1−P(Ac∩Bc) by De Morgan's Law =1−0.3 =0.7 The probability that it will rain today and tomorrow: this is P(A∩B). To find this we note that P(A∩B) =P(A)+P(B)−P(A∪B) by Example 1.10 =0.6+0.5−0.7 =0.4 The probability that it will rain today but not tomorrow: this is P(A∩Bc). P(A∩Bc) =P(A−B) =P(A)−P(A∩B) by Example 1.10 =0.6−0.4 =0.2 The probability that it either will rain today or tomorrow but not both: this is P(A−B)+P(B−A). We have already found P(A−B)=.2. Similarly, we can find P(B−A): P(B−A) =P(B)−P(B∩A) by Example 1.10 =0.5−0.4 =0.1 Thus, P(A−B)+P(B−A)=0.2+0.1=0.3

If the universal set is given by S={1,2,3,4,5,6}, and A={1,2}, B={2,4,5},C={1,5,6} are three sets, find the following sets: A∪B A∩B A¯¯¯¯ B¯¯¯¯ Check De Morgan's law by finding (A∪B)c and Ac∩Bc. Check the distributive law by finding A∩(B∪C) and (A∩B)∪(A∩C)

A∪B={1,2,4,5}. A∩B={2}. A¯¯¯¯={3,4,5,6} (A¯¯¯¯ consists of elements that are in S but not in A). B¯¯¯¯={1,3,6}. We have (A∪B)c={1,2,4,5}c={3,6}, which is the same as Ac∩Bc={3,4,5,6}∩{1,3,6}={3,6}. We have A∩(B∪C)={1,2}∩{1,2,4,5,6}={1,2}, which is the same as (A∩B)∪(A∩C)={2}∪{1}={1,2}.

In a presidential election, there are four candidates. Call them A, B, C, and D. Based on our polling analysis, we estimate that A has a 20 percent chance of winning the election, while B has a 40 percent chance of winning. What is the probability that A or B win the election?

Notice that the events that {A wins}, {B wins}, {C wins}, and {D wins} are disjoint since more than one of them cannot occur at the same time. For example, if A wins, then B cannot win. From the third axiom of probability, the probability of the union of two disjoint events is the summation of individual probabilities. Therefore, P(A wins or B wins) =P({A wins}∪{B wins}) =P({A wins})+P({B wins}) =0.2+0.4 =0.6

I roll a fair die twice and obtain two numbers: X1= result of the first roll, and X2= result of the second roll. Write down the sample space S, and assuming that all outcomes are equally likely (because the die is fair), find the probability of the event A defined as the event that X1+X2=8

P(A)=|A|/|S|=5/36

You roll a fair die. What is the probability of E={1,5}?

P(E)=P({1,5})=P({1})+P({5})=26=13

Using the axioms of probability, prove the following: For any event A, P(Ac)=1−P(A). The probability of the empty set is zero, i.e., P(∅)=0. For any event A, P(A)≤1. P(A−B)=P(A)−P(A∩B). P(A∪B)=P(A)+P(B)−P(A∩B), (inclusion-exclusion principle for n=2). If A⊂B then P(A)≤P(B)

This states that the probability that A does not occur is 1−P(A). To prove it using the axioms, we can write 1 =P(S) (axiom 2) =P(A∪Ac) (definition of compliment) =P(A)+P(Ac) (since A and Ac are disjoint) Since ∅=Sc, we can use part (a) to see that P(∅)=1−P(S)=0. Note that this makes sense as by definition: an event happens if the outcome of the random experiment belongs to that event. Since the empty set does not have any element, the outcome of the experiment never belongs to the empty set. From part (a), P(A)=1−P(Ac) and since P(Ac)≥0 (the first axiom), we have P(A)≤1. We show that P(A)=P(A∩B)+P(A−B). Note that the two sets A∩B and A−B are disjoint and their union is A (Figure 1.17). Thus, by the third axiom of probability P(A)=P((A∩B)∪(A−B))=P(A∩B)+P(A−B)( since A=(A∩B)∪(A−B)) (since A∩B and A−B are disjoint) Note that A and B−A are disjoint sets and their union is A∪B. Thus, P(A∪B) =P(A∪(B−A)) (A∪B=A∪(B−A)) =P(A)+P(B−A) (since A and B−A are disjoint) =P(A)+P(B)−P(A∩B) (by part (d)) Note that A⊂B means that whenever A occurs B occurs, too. Thus intuitively we expect that P(A)≤P(B). Again the proof is similar as before. If A⊂B, then A∩B=A. Thus, P(B) =P(A∩B)+P(B−A) (by part (d)) =P(A)+P(B−A) (since A=A∩B) ≥P(A) (by axiom 1)

In a party, there are 10 people with white shirts and 8 people with red shirts; 4 people have black shoes and white shirts; 3 people have black shoes and red shirts; the total number of people with white or red shirts or black shoes is 21. How many people have black shoes?

|B|=10