The probability of a successful optical alignment inthe assembly of an optical data storage product is 0.7. Assumethe trials are independent. What is the probability that the first two successful alignments require exactly 4 trials?

Given that the probability of a successful optical alignment is \( p = 0.7 \) and the trials are independent, this problem follows a **negative binomial distribution**.

In the negative binomial distribution, we are interested in the probability that a certain number of successful trials (here, 2 successful alignments) occur after a given number of trials (here, 4 trials).

### Problem Breakdown:
- You need exactly 4 trials to get the first 2 successful alignments.
- The probability of success in each trial is \( p = 0.7 \).
- The probability of failure in each trial is \( q = 1 - p = 0.3 \).

The negative binomial formula to find the probability of getting \( k \) successes in \( n \) trials (where the final trial must be a success) is:

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

Where:
- \( n = 4 \) (the total number of trials),
- \( k = 2 \) (the number of successes),
- \( \binom{n-1}{k-1} \) is the binomial coefficient.

### Step-by-Step Solution:
1. The number of trials \( n = 4 \) and the number of successes \( k = 2 \).
2. The binomial coefficient is \( \binom{n-1}{k-1} = \binom{4-1}{2-1} = \binom{3}{1} = 3 \).
3. The probability of success in exactly 2 of these 4 trials is \( p^k = 0.7^2 = 0.49 \).
4. The probability of failure in the remaining \( n-k = 4 - 2 = 2 \) trials is \( q^{n-k} = 0.3^2 = 0.09 \).

Now, calculate the total probability:

\[
P(X = 4) = \binom{3}{1} \cdot (0.7)^2 \cdot (0.3)^2 = 3 \cdot 0.49 \cdot 0.09
\]

\[
P(X = 4) = 3 \cdot 0.0441 = 0.1323
\]

### Final Answer:
The probability that the first two successful alignments require exactly 4 trials is \( 0.1323 \).

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Do you want further details or have any questions?

Here are five related questions to expand the understanding:
1. How would the probability change if the probability of success was 0.6 instead of 0.7?
2. What is the probability that the first three successful alignments require exactly 5 trials?
3. How does the negative binomial distribution differ from the binomial distribution?
4. How do you interpret independence of trials in probability theory?
5. What is the expected number of trials required for the first two successful alignments?

**Tip**: In a negative binomial distribution, the trials stop after a specified number of successes, unlike a binomial distribution, which has a fixed number of trials.

The probability of a successful optical alignment in the assembly of an optical data storage product is 0.7. Assume the trials are independent. What is the probability that the first two successful alignments require exactly 4 trials?

Find out how to calculate the probability of exactly 4 trials to get two successful optical alignments, given a success probability of 0.7. This detailed solution explores the use of the negative binomial distribution in probability, ideal for Grades 11-12 or college-level students.

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