Math Problem Statement

In a product quality inspection, products are tested in two rounds. A product is qualified if it has a rating of 1 or 2, and it is not qualified if it has a rating of 0. The following is the joint probability distribution: 1. If a product is randomly chosen, find the probability that this product is evaluated as qualified in both rounds. 2. If a product is randomly chosen, find the probability that this product is evaluated as qualified in the second round given that it is not qualified in the first round. 3. Compute E(XY).

Solution

Let's solve the problem step by step based on the table and the questions given.

Problem Interpretation:

We are given the joint probability distribution for a product's evaluation in two rounds (Round 1: XX, Round 2: YY), where the product is evaluated with ratings of 0, 1, or 2. The product is qualified if it has a rating of 1 or 2, and not qualified if the rating is 0.

Joint Probability Distribution:

The table provides the joint probabilities for XX and YY, where XX is the rating in Round 1 and YY is the rating in Round 2:

Round 1 (X) \ Round 2 (Y)012
00.1390.160.15
10.2990.1080.041
20.0540.0140.035

1. Find the probability that the product is evaluated as qualified in both rounds.

The product is qualified in both rounds if X{1,2}X \in \{1, 2\} and Y{1,2}Y \in \{1, 2\}. We sum the joint probabilities where both XX and YY are either 1 or 2:

P(Qualified in both rounds)=P(X=1,Y=1)+P(X=1,Y=2)+P(X=2,Y=1)+P(X=2,Y=2)P(\text{Qualified in both rounds}) = P(X=1, Y=1) + P(X=1, Y=2) + P(X=2, Y=1) + P(X=2, Y=2) From the table:

P(Qualified in both rounds)=0.108+0.041+0.014+0.035=0.198P(\text{Qualified in both rounds}) = 0.108 + 0.041 + 0.014 + 0.035 = 0.198

2. Find the probability that the product is qualified in the second round given that it is not qualified in the first round.

The product is not qualified in the first round if X=0X = 0, and it is qualified in the second round if Y{1,2}Y \in \{1, 2\}.

We are looking for: P(Y{1,2}X=0)=P(X=0,Y{1,2})P(X=0)P(Y \in \{1, 2\} \mid X = 0) = \frac{P(X=0, Y \in \{1, 2\})}{P(X=0)}

Where:

  • P(X=0,Y{1,2})=P(X=0,Y=1)+P(X=0,Y=2)=0.16+0.15=0.31P(X=0, Y \in \{1, 2\}) = P(X=0, Y=1) + P(X=0, Y=2) = 0.16 + 0.15 = 0.31
  • P(X=0)=P(X=0,Y=0)+P(X=0,Y=1)+P(X=0,Y=2)=0.139+0.16+0.15=0.449P(X=0) = P(X=0, Y=0) + P(X=0, Y=1) + P(X=0, Y=2) = 0.139 + 0.16 + 0.15 = 0.449

Thus:

P(Y{1,2}X=0)=0.310.4490.69P(Y \in \{1, 2\} \mid X=0) = \frac{0.31}{0.449} \approx 0.69

3. Compute E(XY)E(XY).

The expected value E(XY)E(XY) is calculated as the sum of X×Y×P(X,Y)X \times Y \times P(X, Y) for all possible values of XX and YY:

E(XY)=X,YX×Y×P(X,Y)E(XY) = \sum_{X, Y} X \times Y \times P(X, Y)

We calculate each term individually using the probabilities from the table:

  • E(XY)=(0×0×0.139)+(0×1×0.16)+(0×2×0.15)E(XY) = (0 \times 0 \times 0.139) + (0 \times 1 \times 0.16) + (0 \times 2 \times 0.15)
  • +(1×0×0.299)+(1×1×0.108)+(1×2×0.041)+ (1 \times 0 \times 0.299) + (1 \times 1 \times 0.108) + (1 \times 2 \times 0.041)
  • +(2×0×0.054)+(2×1×0.014)+(2×2×0.035)+ (2 \times 0 \times 0.054) + (2 \times 1 \times 0.014) + (2 \times 2 \times 0.035)

Simplifying:

E(XY)=0+0+0+0+0.108+0.082+0+0.028+0.14=0.358E(XY) = 0 + 0 + 0 + 0 + 0.108 + 0.082 + 0 + 0.028 + 0.14 = 0.358


Summary of Answers:

  1. Probability that the product is evaluated as qualified in both rounds: 0.198.
  2. Probability that the product is qualified in the second round given that it is not qualified in the first round: 0.69.
  3. E(XY)E(XY): 0.358.

Would you like any further details or clarifications?

Here are some related questions:

  1. How would the probabilities change if a new joint probability table was given?
  2. What if the qualifications were based on different thresholds?
  3. How would you compute E(X2)E(X^2) or E(Y2)E(Y^2) based on the same table?
  4. How does conditional probability work in more complex joint distributions?
  5. Can we visualize the probabilities using charts or graphs?

Tip: Always verify that the probabilities in a joint distribution sum to 1 as a consistency check!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Probability
Joint Probability Distribution
Conditional Probability
Expected Value

Formulas

P(Qualified in both rounds) = P(X=1, Y=1) + P(X=1, Y=2) + P(X=2, Y=1) + P(X=2, Y=2)
Conditional Probability: P(B|A) = P(A and B) / P(A)
Expected Value: E(XY) = Σ(X * Y * P(X, Y))

Theorems

Law of Total Probability
Definition of Expected Value

Suitable Grade Level

Grades 11-12, College Level