Question

1 is \(0.8\). If you get contract #1, the probability you also get contract #… # 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

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

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.

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.

Calculate Probability of Both Contracts

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

Probability of Getting No Contract

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

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, \( P(X=0) = 0.14 \), \( P(X=1) = 0.70 \), \( P(X=2) = 0.16 \).

Calculate Expected Value

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

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 \( \sqrt{\text{Var}(X)} = \sqrt{0.2996} \approx 0.547 \).

Key concepts

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.