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Chapter 16: Problem 22

1 is \(0.8\). If you get contract #1, the probability you also get contract #鈥� # Contracts. Your company bids for two contracts. You believe the probability you get contract #1 is \(0.8\). If you get contract #1, the probability you also get contract #2 will be \(0.2\), and if you do not get \(\\# 1\), the probability you get #2 will be \(0.3\). a) Are the two contracts independent? Explain. b) Find the probability you get both contracts. c) Find the probability you get no contract. d) Let \(X\) be the number of contracts you get. Find the probability model for \(X\). e) Find the expected value and standard deviation.

Short Answer

Expert verified
a) Independent? No. b) Probability both: 0.16. c) Probability none: 0.14. d) X: 0.14, 0.70, 0.16. e) E(X) = 1.02, Std Dev \( \approx 0.547 \).

Step by step solution

01

Understand Independence

Two events are independent if the probability of both events occurring is the product of their individual probabilities. If the probability of getting contract #2 given contract #1, \( P(C_2|C_1) \), equals the probability of getting contract #2, \( P(C_2) \), then the events are independent.
02

Check Contract Independence

Calculate \( P(C_2) \): \( P(C_2) = P(C_1) \times P(C_2 | C_1) + P(C_1^c) \times P(C_2 | C_1^c) \). Substitute the given values: \(0.8 \times 0.2 + 0.2 \times 0.3 = 0.16 + 0.06 = 0.22\). Since \( P(C_2 | C_1) = 0.2 \) and \( P(C_2) = 0.22 \), they are not equal, hence the contracts are not independent.
03

Calculate Probability of Both Contracts

The probability you get both contracts is \( P(C_1 \cap C_2) = P(C_1) \times P(C_2 | C_1) = 0.8 \times 0.2 = 0.16 \).
04

Probability of Getting No Contract

The probability of getting no contract is \( P(C_1^c \cap C_2^c) = P(C_1^c) \times P(C_2^c | C_1^c) = 0.2 \times (1-0.3) = 0.2 \times 0.7 = 0.14 \).
05

Probability Model for X

Define \( X \) as the number of contracts won. Possible values are 0, 1, and 2. Calculate probabilities:- \( P(X=0) = 0.14 \), already found.- \( P(X=1) \) means only either contract #1 or #2 is won: \( P(C_1 \cap C_2^c) + P(C_1^c \cap C_2) = 0.8 \times 0.8 + 0.2 \times 0.3 = 0.64 + 0.06 = 0.70 \).- \( P(X=2) = P(C_1 \cap C_2) = 0.16 \).Thus, \( P(X=0) = 0.14 \), \( P(X=1) = 0.70 \), \( P(X=2) = 0.16 \).
06

Calculate Expected Value

The expected value \( E(X) \) is calculated as follows: \( E(X) = 0 \cdot P(X=0) + 1 \cdot P(X=1) + 2 \cdot P(X=2) = 0 + 0.70 + 0.32 = 1.02 \).
07

Calculate Standard Deviation

First, find \( E(X^2) \): \( E(X^2) = 0^2 \cdot P(X=0) + 1^2 \cdot P(X=1) + 2^2 \cdot P(X=2) = 0 + 0.70 + 0.64 = 1.34 \). Then, use the variance formula \( \text{Var}(X) = E(X^2) - (E(X))^2 = 1.34 - (1.02)^2 = 1.34 - 1.0404 = 0.2996 \). The standard deviation is \( \sqrt{\text{Var}(X)} = \sqrt{0.2996} \approx 0.547 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Independence of Events
Two events are considered independent when the occurrence of one event does not affect the probability of the other event occurring. In simpler terms, if knowing the result of one event gives you no information about the result of the second event, these events are independent.
For events A and B, if they are independent, the probability of both occurring is the product of their probabilities:
  • \( P(A \cap B) = P(A) \times P(B) \)
If we refer back to the bidding scenario in the exercise, independence checks whether the outcome of the first contract affects the outcome of the second. Here, the probabilities provided for winning the second contract differ based on the outcome of the first. Thus, the contracts are not independent, as having contract #1 changes the likelihood of winning contract #2.
Expected Value
The expected value is key in decision-making, as it provides a measure of the center of a probability distribution. It is essentially the long-term average result, over many repetitions, of a random experiment. To calculate the expected value, multiply each possible outcome by its probability, then sum up the results. This gives you a weighted average of all possible outcomes.
In our scenario, we define \( X \) as the number of contracts won, with possible values of 0, 1, or 2. To find the expected value, calculate:
  • \( E(X) = 0 \cdot P(X=0) + 1 \cdot P(X=1) + 2 \cdot P(X=2) \)
The probabilities from the exercise tell us that these values are \( P(X=0) = 0.14 \), \( P(X=1) = 0.70 \), and \( P(X=2) = 0.16 \). Calculating this gives you an expected value of approximately 1.02 contracts won.
Standard Deviation
The standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation means the values are close to the expected value, while a high standard deviation indicates there are values that are spread out over a wider range. It is an essential tool in probability theory, as it helps to understand the reliability of the expected value.
For our exercise, the standard deviation can be found using the formula:
  • \( \text{Std Dev}(X) = \sqrt{\text{Var}(X)} \)
  • Where the variance \( \text{Var}(X) = E(X^2) - (E(X))^2 \)
From earlier calculations, \( E(X^2) \) was determined to be 1.34, and \( E(X) = 1.02 \). Consequently, the variance is \( 0.2996 \), leading to a standard deviation calculation of approximately 0.547, indicating a moderate spread around the expected value of contracts won.

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Most popular questions from this chapter

Pick a card, any card. You draw a card from a deck. If you get a red card, you win nothing. If you get a spade, you win \(\$ 5\). For any club, you win \(\$ 10\) plus an extra \(\$ 20\) for the ace of clubs. a) Create a probability model for the amount you win. b) Find the expected amount you'll win. c) What would you be willing to pay to play this game?

Cereal. The amount of cereal that can be poured into a small bowl varies with a mean of \(1.5\) ounces and a standard deviation of \(0.3\) ounces. A large bowl holds a mean of \(2.5\) ounces with a standard deviation of \(0.4\) ounces. You open a new box of cereal and pour one large and one small bowl. a) How much more cereal do you expect to be in the large bowl? b) What's the standard deviation of this difference? c) If the difference follows a Normal model, what's the probability the small bowl contains more cereal than the large one? d) What are the mean and standard deviation of the total amount of cereal in the two bowls? e) If the total follows a Normal model, what's the probability you poured out more than \(4.5\) ounces of cereal in the two bowls together? f) The amount of cereal the manufacturer puts in the boxes is a random variable with a mean of \(16.3\) ounces and a standard deviation of \(0.2\) ounces. Find the expected amount of cereal left in the box and the standard deviation.

Weightlifting. The Atlas BodyBuilding Company (ABC) sells "starter sets" of barbells that consist of one bar, two 20-pound weights, and four 5 -pound weights. The bars weigh an average of 10 pounds with a standard deviation of \(0.25\) pounds. The weights average the specified amounts, but the standard deviations are \(0.2\) pounds for the 20-pounders and \(0.1\) pounds for the 5 -pounders. We can assume that all the weights are normally distributed. a) \(\mathrm{ABC}\) ships these starter sets to customers in two boxes: The bar goes in one box and the six weights go in another. What's the probability that the total weight in that second box exceeds \(60.5\) pounds? Define your variables clearly and state any assumptions you make. b) It costs \(\mathrm{ABC} \$ 0.40\) per pound to ship the box containing the weights. Because it's an odd-shaped package, though, shipping the bar costs \(\$ 0.50\) a pound plus a \(\$ 6.00\) surcharge. Find the mean and standard deviation of the company's total cost for shipping a starter set. c) Suppose a customer puts a 20-pound weight at one end of the bar and the four 5 -pound weights at the other end. Although he expects the two ends to weigh the same, they might differ slightly. What's the probability the difference is more than a quarter of a pound?

Batteries. In a group of 10 batteries, 3 are dead. You choose 2 batteries at random. a) Create a probability model for the number of good batteries you get. b) What's the expected number of good ones you get? c) What's the standard deviation?

Defects. A consumer organization inspecting new cars found that many had appearance defects (dents, scratches, paint chips, etc.). While none had more than three of these defects, \(7 \%\) had three, \(11 \%\) two, and \(21 \%\) one defect. Find the expected number of appearance defects in a new car and the standard deviation.
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