To solve these probability questions, we can use the binomial probability distribution since we're dealing with a fixed number of trials (10 light bulbs), two possible outcomes (defective or not defective), and a known probability of a defective light bulb (19%).

Formula for Binomial Probability:

The probability of exactly $k$ defective bulbs in $n$ trials (with probability $p$ of success for each trial) is given by the binomial probability formula:

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

Where:

• $n = 10$ (number of trials)
• $p = 0.19$ (probability of failure or defective)
• $k$ is the number of defective bulbs we're interested in
• $\binom{n}{k}$ is the binomial coefficient, which calculates the number of ways to choose $k$ defective bulbs out of $n$

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

For $k = 0$, the probability is:

$P(X = 0) = \binom{10}{0} (0.19)^0 (0.81)^{10}$ $P(X = 0) = 1 \times 1 \times (0.81)^{10} \approx 0.1216$

So, the probability that none of the light bulbs are defective is approximately 0.1216.

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

For $k = 1$, the probability is:

$P(X = 1) = \binom{10}{1} (0.19)^1 (0.81)^9$ $P(X = 1) = 10 \times 0.19 \times (0.81)^9 \approx 0.2852$

So, the probability that exactly one of the light bulbs is defective is approximately 0.2852.

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

For $k \leq 3$, we need to sum the probabilities of having 0, 1, 2, and 3 defective bulbs:

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

We already know the probabilities for $P(X = 0)$ and $P(X = 1)$. Now we need to calculate $P(X = 2)$ and $P(X = 3)$.

For $P(X = 2)$:

$P(X = 2) = \binom{10}{2} (0.19)^2 (0.81)^8 \approx 45 \times (0.19)^2 \times (0.81)^8 \approx 0.2920$

For $P(X = 3)$:

$P(X = 3) = \binom{10}{3} (0.19)^3 (0.81)^7 \approx 120 \times (0.19)^3 \times (0.81)^7 \approx 0.1901$

Now sum all of these values:

$P(X \leq 3) = 0.1216 + 0.2852 + 0.2920 + 0.1901 \approx 0.8889$

So, the probability that three or fewer of the LED light bulbs are defective is approximately 0.8889.

Final Answers:

• (a) The probability that none of the light bulbs are defective: 0.1216
• (b) The probability that exactly one light bulb is defective: 0.2852
• (c) The probability that three or fewer of the light bulbs are defective: 0.8889

Let me know if you'd like to explore more about binomial distributions or need further clarification!

Related Questions:

1. How would the probabilities change if the failure rate were 15% instead of 19%?
2. What does the binomial coefficient $\binom{n}{k}$ represent in the context of this problem?
3. How can you calculate the probability of more than 3 defective bulbs?
4. How would the probabilities change if the sample size was increased to 15 bulbs?
5. How does the binomial distribution relate to the normal distribution in large samples?

Tip: When working with binomial probabilities, using a calculator or statistical software can help you quickly compute binomial coefficients and powers, especially for large $n$ or $k$.