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 2
​%. 
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 enter your response here
.
​(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 enter your response here
.
​(Type an integer or a decimal. Round to four decimal places as​ needed.)
Part 3
c. What is the probability that two
 
or fewer of the LED light bulbs are​ defective?
The probability that two
 
or fewer of the LED light bulbs are defective is enter your response here
.
​(Type an integer or a decimal. Round to four decimal places as​ needed.)
Part 4
d. What is the probability that three
 
or more of the LED light bulbs are​ defective?
The probability that three
 
or more of the LED light bulbs are defective is enter your response here
.
​(Type an integer or a decimal. Round to four decimal places as​ needed.)

To solve this problem, we use the **binomial probability formula** since we have a fixed number of trials, each with two possible outcomes (defective or not defective). The binomial formula is:

\[
P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}
\]

Where:
- \( n \) = number of trials (in this case, 10 LED light bulbs)
- \( k \) = number of successes (defective light bulbs)
- \( p \) = probability of success (defective bulb, which is 2% or 0.02)
- \( 1 - p \) = probability of failure (not defective, which is 98% or 0.98)

Let's solve each part:

---

### Part 1: Probability that none of the LED light bulbs are defective

This means \( k = 0 \). The formula becomes:

\[
P(X = 0) = \binom{10}{0} (0.02)^0 (0.98)^{10} = (0.98)^{10}
\]

Calculating this:

\[
P(X = 0) = (0.98)^{10} \approx 0.8171
\]

So, the probability that none of the LED light bulbs are defective is approximately **0.8171**.

---

### Part 2: Probability that exactly one of the LED light bulbs is defective

Here, \( k = 1 \). The formula becomes:

\[
P(X = 1) = \binom{10}{1} (0.02)^1 (0.98)^9
\]

Calculating this:

\[
\binom{10}{1} = 10
\]

\[
P(X = 1) = 10 \times (0.02) \times (0.98)^9 \approx 0.1665
\]

So, the probability that exactly one of the LED light bulbs is defective is approximately **0.1665**.

---

### Part 3: Probability that two or fewer of the LED light bulbs are defective

This means finding \( P(X \leq 2) \), which is the sum of \( P(X = 0) \), \( P(X = 1) \), and \( P(X = 2) \):

\[
P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)
\]

We already have \( P(X = 0) \approx 0.8171 \) and \( P(X = 1) \approx 0.1665 \).

Now, let's find \( P(X = 2) \):

\[
P(X = 2) = \binom{10}{2} (0.02)^2 (0.98)^8
\]

Calculating this:

\[
\binom{10}{2} = 45
\]

\[
P(X = 2) = 45 \times (0.02)^2 \times (0.98)^8 \approx 0.0149
\]

Now, add them up:

\[
P(X \leq 2) \approx 0.8171 + 0.1665 + 0.0149 \approx 0.9985
\]

So, the probability that two or fewer of the LED light bulbs are defective is approximately **0.9985**.

---

### Part 4: Probability that three or more of the LED light bulbs are defective

This is the complement of \( P(X \leq 2) \):

\[
P(X \geq 3) = 1 - P(X \leq 2)
\]

Using the value we calculated for \( P(X \leq 2) \):

\[
P(X \geq 3) = 1 - 0.9985 \approx 0.0015
\]

So, the probability that three or more of the LED light bulbs are defective is approximately **0.0015**.

---

### Summary:
- a. Probability that none are defective: **0.8171**
- b. Probability that exactly one is defective: **0.1665**
- c. Probability that two or fewer are defective: **0.9985**
- d. Probability that three or more are defective: **0.0015**

Would you like further details on any of these calculations or need help with related concepts? Here are some related questions to explore:

1. What if the sample size changes to 15 bulbs? How would the probabilities change?
2. How do we calculate the mean and variance of a binomial distribution?
3. How would the probability change if the failure rate increases to 5%?
4. What is the expected number of defective bulbs in a sample of 10?
5. How would the probabilities change if we selected a sample of 20 bulbs?

**Tip:** The sum of probabilities for all possible outcomes in a probability distribution is always 1. This principle helps verify calculations like in part (d).

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 2%. Suppose a random sample of 10 LED light bulbs is selected. a. What is the probability that none of the LED light bulbs are defective? b. What is the probability that exactly one of the LED light bulbs is defective? c. What is the probability that two or fewer of the LED light bulbs are defective? d. What is the probability that three or more of the LED light bulbs are defective?

Learn how to calculate probabilities using binomial distribution for LED light bulb defects. With a 2% defect rate, solve questions for none, one, two, or more defective bulbs in a sample of 10. Includes step-by-step calculations using binomial formulas.

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