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:

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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.

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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.

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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.

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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.

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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).