Try It Now
Site: | Saylor Academy |
Course: | CS202: Discrete Structures |
Book: | Try It Now |
Printed by: | Guest user |
Date: | Thursday, 3 April 2025, 6:38 PM |
Description
Work these exercises to see how well you understand this material.
Exercises
- For two events A and B, P(A) = 0.73, P(B) = 0.48, and P(A ∩ B) = 0.29.
- Find P(A|B).
- Find P(B|A).
- Determine whether or not A and B are independent.
- For independent events A and B, P(A) = 0.81 and P(B) = 0.27.
- Find P(A ∩ B).
- Find P(A|B).
- Find P(B|A).
- For mutually exclusive events A and B, P(A) = 0.17 and P(B) = 0.32.
- Find P(A|B).
- Find P(B|A).
- Compute the following probabilities in connection with the roll of a single fair die.
- The probability that the roll is even.
- The probability that the roll is even, given that it is not a two.
- The probability that the roll is even, given that it is not a one.
- A special deck of 16 cards has 4 that are blue, 4 yellow, 4 green, and 4 red. The four cards of each color are numbered from one to four. A single card is drawn at random. Find the following probabilities.
- The probability that the card drawn is red.
- The probability that the card is red, given that it is not green.
- The probability that the card is red, given that it is neither red nor yellow.
- The probability that the card is red, given that it is not a four.
- A random experiment gave rise to the two-way contingency table shown. Use it to compute the probabilities indicated.
R S A 0.13 0.07 B 0.61 0.19
- P(A), P(R), P(A ∩ R).
- Based on the answer to (a), determine whether or not the events A and R are independent.
- Based on the answer to (b), determine whether or not P(A|R) can be predicted without any computation. If so, make the prediction. In any case, compute P(A|R) using the Rule for Conditional Probability.
- Suppose for events A and B in a random experiment P(A) = 0.70 and P(B) = 0.30. Compute the indicated probability, or explain why there is not enough information to do so.
- P(A ∩ B).
- P(A ∩ B), with the extra information that A and B are independent.
- P(A ∩ B), with the extra information that A and B are mutually exclusive.
- Suppose for events A, B, and C connected to some random experiment, A, B, and C are independent and P(A) = 0.88, P(B) = 0.65, and P(C) = 0.44. Compute the indicated probability, or explain why there is not enough information to do so.
- P(A ∩ B ∩ C)
- P(Ac ∩ Bc ∩ Cc)
- The sample space that describes all three-child families according to the genders of the children with respect to birth order is:
S = {bbb, bbg, bgb, bgg, gbb, gbg, ggb, ggg}
In the experiment of selecting a three-child family at random, compute each of the following probabilities, assuming all outcomes are equally likely.- The probability that the family has at least two boys.
- The probability that the family has at least two boys, given that not all of the children are girls.
- The probability that at least one child is a boy.
- The probability that at least one child is a boy, given that the first born is a girl.
- The following two-way contingency table gives the breakdown of the population in a particular locale according to party affiliation (A, B, C, or None) and opinion on a bond issue:
Affiliation Opinion Favors Opposes Undecided A 0.12 0.09 0.07 B 0.16 0.12 0.14 C 0.04 0.03 0.06 None 0.08 0.06 0.03
A person is selected at random. Find each of the following probabilities.- The person is in favor of the bond issue.
- The person is in favor of the bond issue, given that he is affiliated with party A.
- The person is in favor of the bond issue, given that he is affiliated with party B.
- The following two-way contingency table gives the breakdown of the population of adults in a particular locale according to employment type and level of life insurance:
Employment Type Level of Insurance Low Medium High Unskilled 0.07 0.19 0.00 Semi-skilled 0.04 0.28 0.08 Skilled 0.03 0.18 0.05 Professional 0.01 0.05 0.02
An adult is selected at random. Find each of the following probabilities.- The person has a high level of life insurance.
- The person has a high level of life insurance, given that he does not have a professional position.
- The person has a high level of life insurance, given that he has a professional position.
- Determine whether or not the events "has a high level of life insurance" and "has a professional position" are independent.
- The sensitivity of a drug test is the probability that the test will be positive when administered to a person who has actually taken the drug. Suppose that there are two independent tests to detect the presence of a certain type of banned drugs in athletes. One has sensitivity 0.75; the other has sensitivity 0.85. If both are applied to an athlete who has taken this type of drug, what is the chance that his usage will go undetected?
- An accountant has observed that 5% of all copies of a particular two-part form have an error in Part I, and 2% have an error in Part II. If the errors occur independently, find the probability that a randomly selected form will be error-free.
- Events A and B are mutually exclusive. Find P(A|B).
- A basketball player makes 60% of the free throws that he attempts, except that if he has just tried and missed a free throw then his chances of making a second one go down to only 30%. Suppose he has just been awarded two free throws.
- Find the probability that he makes both.
- Find the probability that he makes at least one. (A tree diagram could help.)
Source: https://saylordotorg.github.io/text_introductory-statistics/index.html This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License.
Solutions
-
- 0.6
- 0.4
- not independent
-
- 0.22
- 0.81
- 0.27
-
- 0
- 0
-
- 0.5
- 0.4
- 0.6
-
- 0.25
- 0.33
- 0
- 0.25
-
- P(A) = 0.3, P(R) = 0.4, P(A ∩ R) = 0.12
- independent
- without computation 0.3
-
- Insufficient information. The events A and B are not known to be either independent or mutually exclusive.
- 0.21
- 0
-
- 0.25
- 0.02
-
- 0.5
- 0.57
- 0.875
- 0.75
-
- 0.4
- 0.43
- 0.38
-
- 0.15
- 0.14
- 0.25
- not independent
- 0.0375
- 0.931
- 0
-
- 0.36
- 0.72