Math Problem Statement

A manufacturing company regularly conducts quality control checks at specified periods on the products it manufactures.​ Historically, the failure rate for LED light bulbs that the company manufactures is 13​%. Suppose a random sample of 10 LED light bulbs is selected. Complete parts​ (a) through​ (d) below. Question content area bottom Part 1 a. What is the probability that none of the LED light bulbs are​ defective? The probability that none of the LED light bulbs are defective is    0.2484. ​(Type an integer or a decimal. Round to four decimal places as​ needed.) Part 2 b. What is the probability that exactly one of the LED light bulbs is​ defective? The probability that exactly one of the LED light bulbs is defective is

Solution

To solve this, we’ll use the binomial probability formula, which applies here since each LED light bulb has two outcomes (defective or not defective) and the probability of failure is constant for each bulb.

The binomial probability formula is:

P(X=k)=(nk)pk(1p)nkP(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}

where:

  • n=10n = 10 (the number of trials or light bulbs in this case),
  • kk is the number of defective bulbs,
  • p=0.13p = 0.13 (the probability of a bulb being defective),
  • 1p=0.871 - p = 0.87 (the probability of a bulb not being defective).

Part (a)

We need to find the probability that none of the light bulbs are defective, so k=0k = 0.

P(X=0)=(100)(0.13)0(0.87)10P(X = 0) = \binom{10}{0} (0.13)^0 (0.87)^{10}

Calculating this:

P(X=0)=1(0.87)100.2484P(X = 0) = 1 \cdot (0.87)^{10} \approx 0.2484

This matches the provided answer.

Part (b)

Now, let's find the probability that exactly one of the LED light bulbs is defective, so k=1k = 1.

P(X=1)=(101)(0.13)1(0.87)9P(X = 1) = \binom{10}{1} (0.13)^1 (0.87)^9

  1. Calculate (101)=10\binom{10}{1} = 10 (since there are 10 ways to choose one defective bulb out of 10).
  2. Then, calculate (0.13)1=0.13(0.13)^1 = 0.13.
  3. Next, calculate (0.87)90.4228(0.87)^9 \approx 0.4228.
  4. Finally, multiply these values:

P(X=1)=100.130.42280.5496P(X = 1) = 10 \cdot 0.13 \cdot 0.4228 \approx 0.5496

So, the probability that exactly one of the LED light bulbs is defective is approximately 0.5496.

Would you like further details, or do you have any questions about these steps?


Here are five related questions to deepen your understanding:

  1. How would you calculate the probability that exactly two of the LED light bulbs are defective?
  2. What is the probability that at least one of the LED light bulbs is defective?
  3. How would the results change if the sample size were increased to 15 light bulbs?
  4. What is the expected number of defective light bulbs in a sample of 10?
  5. How does changing the failure rate affect the probabilities calculated here?

Tip: The binomial probability formula is useful for any scenario with two possible outcomes (e.g., defective vs. not defective) and a constant probability of success or failure.

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Math Problem Analysis

Mathematical Concepts

Probability
Binomial Distribution

Formulas

Binomial probability formula: P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)

Theorems

Binomial Theorem

Suitable Grade Level

College-level or Advanced High School