Example Problems on Set Theory and Probability
ST701 Homework 1 on Set Theory and Probability
Problems: 1.1, 1.4, 1.5, 1.13, 1.23, Problem 6
1.1
For each of the following experiments, describe the sample space.
a.
Toss a coin four times.
This will look like a sequence of 4 heads or tails. e.g HHHH,HHHT,HHTH,…TTTT. There will be 16 elements.
b.
Count the number of insect damaged leaves on a plant.
This will be the natural numbers, [0,∞), assuming there is no limit to the possible number of leaves on a plant.
c.
Measure the lifetime (in hours) of a particular brand of light bulb.
This can be any real number greater than or equal to 0, [0,∞).
d.
Record the weights of 10-day-old rats.
This can be any real number greater than 0, (0,∞).
e.
Observe the proportion of defectives in a shipment of electronic components.
This can be any real number between 0 and 1, [0,1].
1.4
For events A and B, find formulas for the probabilities of the following events in terms of quantities P(A), P(B), and P(A∩B).
a.
either in A or B or both
P(A∪B)=P(A)+P(B)−P(A∩B)by theorem 3.2b.
either A or B but not both
P((A∪B)−(A∩B))=P(A)+P(B)−2P(A∩B)c.
at least one of A or B
P(A∪B)=P(A)+P(B)−P(A∩B)by theorem 3.2d.
at most one of A or B
1−P(A∩B)1.5
Approximately one-third of all human twins are identical (one-egg) and two-thirds are fraternal (two-egg) twins. Identical twins are necessarily the same sex, with male and female being equally likely. Among fraternal twins, approximately one-fourth are both female, one-fourth are both male, and half are one male and one female. Finally, among all U.S. births, approximately 1 in 90 is a twin birth. Define the following events:
A={ a U.S. birth results in twin females }B= { a U.S. birth results in identical twins } C= { a U.S. birth results in twins. }a.
State in words, the event A∩B∩C.
The set of U.S. birth results in female, identical twins. (Notice that A and B are subsets of C.)
b.
Find P(A∩B∩C).
P(A∩B∩C)=190⋅13⋅12=1540Where
190=The proportion of twin births,13=The proportion of twins that are identical,12=The proportion of identical twins that are female.1.13
If P(A)=13 and P(BC)=14, can A and B be disjoint? Explain.
No, A and B cannot be disjoint. Notice that P(B)=1−P(BC)=34. If we assume that A and B are disjoint, then A∩B=∅; however P(A)+P(B)=1312>1. Thus, A and B cannot be disjoint and some elements must be in their intersection.
1.23
Two people each toss a fair coin n times. Find the probability that they will toss the same number of heads. (Answer: (\frac{ 1 }{ 4 })^n \cdot {2n \choose n} )
Notice that the process of flipping a coin is a binomial distribution with n trials and a probability of success 1/2. Notice that the outcome of each person is indepent of the other. Since we are looking for when they flip the same number of heads we are essentially looking for P(X=k) \land P(Y=k) (where X and Y represent the outcomes of the first and second person respectively). Since these events are independent, we can multiply their CDFs together.
\begin{align} \sum_{k=1}^{n} \Big[ {n \choose k} \frac{ 1 }{ 2 }^k \frac{ 1 }{ 2 }^{n-k} \Big] \cdot \Big[ {n \choose k} \frac{ 1 }{ 2 }^k \frac{ 1 }{ 2 }^{n-k} \Big] & = \sum_{k=1}^{n} {n \choose k} {n \choose k} \frac{ 1 }{ 2 }^{2n} \\ & = \frac{ 1 }{ 2 }^{2n} \sum_{k=1}^{n} {n \choose k} {n \choose k} \\ & = \frac{ 1 }{ 2 }^{2n} \cdot {2n \choose n} & \text{by Vandermonde's identity} \\ & = \Big( \frac{ 1 }{ 4 } \Big)^n \cdot {2n \choose n} \end{align}6
Prove rigorously that
P(A\cup B\cup C) = P(A) + P(B) + P(C) - P(A\cap B) - P(A\cap C) - P(B\cap C) + P(A\cap B\cap C)(Hint: use P(A\cup B) = P(A) + P(B) - P(A\cap B) and Theorem 1.1.4 in the textbook.) Note Venn diagram is not sufficient for a proof here.
\begin{align} P(A \cup B \cup C) & = P(A \cup (B \cup C)) \\ & = P(A) + P(B \cup C) - P(A \cap (B \cup C)) \\ & = P(A) + P(B) + P(C) - P(B \cap C) - P(A \cap (B \cup C)) \\ & = P(A) + P(B) + P(C) - P(B \cap C) - P( (A \cap B) \cup (A \cap C)) \\ & = P(A) + P(B) + P(C) - P(B \cap C) - (P(A \cap B) + P( A \cap C) - P(A \cap B \cap C)) \\ & = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C) \end{align}We can check this with a Venn diagram.