Math Problem Statement

Suppose that a box contains 6 cameras and that 5 of them are defective. A sample of 2 cameras is selected at random, with replacement. Define the random variable as the number of defective cameras in the sample.

Write the probability distribution for . P()

What is the expected value of ?
Submit Question Question 7

Solution

To define the random variable XX as the number of defective cameras in a sample of 2 selected at random with replacement from a box containing 6 cameras (5 defective and 1 non-defective), we can first determine the probabilities associated with each possible outcome.

Probability Distribution

The possible values of XX are 0, 1, or 2, representing the number of defective cameras in the sample. Since the selection is done with replacement, the probabilities for each sample can be calculated as follows:

  1. Probability of selecting a defective camera (D): P(D)=56P(D) = \frac{5}{6}

  2. Probability of selecting a non-defective camera (N): P(N)=16P(N) = \frac{1}{6}

Now, we calculate the probabilities for each possible value of XX:

  • P(X=0)P(X = 0): Both selected cameras are non-defective. P(X=0)=P(N)×P(N)=(16)×(16)=136P(X = 0) = P(N) \times P(N) = \left(\frac{1}{6}\right) \times \left(\frac{1}{6}\right) = \frac{1}{36}

  • P(X=1)P(X = 1): One camera is defective and one is non-defective. This can happen in two ways: either the first camera is defective and the second is non-defective, or vice versa. P(X=1)=P(D)×P(N)+P(N)×P(D)=(56)×(16)+(16)×(56)=2×(536)=1036=518P(X = 1) = P(D) \times P(N) + P(N) \times P(D) = \left(\frac{5}{6}\right) \times \left(\frac{1}{6}\right) + \left(\frac{1}{6}\right) \times \left(\frac{5}{6}\right) = 2 \times \left(\frac{5}{36}\right) = \frac{10}{36} = \frac{5}{18}

  • P(X=2)P(X = 2): Both selected cameras are defective. P(X=2)=P(D)×P(D)=(56)×(56)=2536P(X = 2) = P(D) \times P(D) = \left(\frac{5}{6}\right) \times \left(\frac{5}{6}\right) = \frac{25}{36}

Summary of the Probability Distribution

  • P(X=0)=136P(X = 0) = \frac{1}{36}
  • P(X=1)=518P(X = 1) = \frac{5}{18}
  • P(X=2)=2536P(X = 2) = \frac{25}{36}

Expected Value of XX

The expected value E(X)E(X) is calculated using the formula: E(X)=(xP(X=x))E(X) = \sum (x \cdot P(X = x)) Calculating it step-by-step: E(X)=0P(X=0)+1P(X=1)+2P(X=2)E(X) = 0 \cdot P(X = 0) + 1 \cdot P(X = 1) + 2 \cdot P(X = 2) E(X)=0136+1518+22536E(X) = 0 \cdot \frac{1}{36} + 1 \cdot \frac{5}{18} + 2 \cdot \frac{25}{36} E(X)=0+518+5036E(X) = 0 + \frac{5}{18} + \frac{50}{36} Convert 518\frac{5}{18} to have a common denominator with 5036\frac{50}{36}: 518=1036\frac{5}{18} = \frac{10}{36} So, E(X)=1036+5036=6036=53E(X) = \frac{10}{36} + \frac{50}{36} = \frac{60}{36} = \frac{5}{3}

Final Results

  • Probability Distribution:

    • P(X=0)=136P(X = 0) = \frac{1}{36}
    • P(X=1)=518P(X = 1) = \frac{5}{18}
    • P(X=2)=2536P(X = 2) = \frac{25}{36}
  • Expected Value: E(X)=53E(X) = \frac{5}{3}

If you have any further questions or need more details, feel free to ask! Here are some related questions to consider:

  1. How would the probabilities change if there were 4 defective cameras?
  2. What is the variance of the random variable XX?
  3. How would the expected value change if the sample size increased to 3 cameras?
  4. What are the implications of sampling without replacement?
  5. How does the probability distribution affect decision-making in quality control?

Tip: Always check if the selection is with or without replacement, as it significantly impacts the probabilities and calculations!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Probability
Random Variables

Formulas

P(X = x) = (probability of selection) ^ (number of selections)
E(X) = sum(x * P(X = x))

Theorems

Binomial Probability Distribution

Suitable Grade Level

Grades 10-12