To solve these problems, we will use the binomial distribution because the outcomes for each goblet are independent, and the probability of a goblet being a "second" remains constant at 10% for each trial.

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Problem a: Probability that only one is a second among six goblets

The probability of $k$ successes (in this case, defective goblets) in $n$ trials is given by the binomial probability formula:

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

Where:

• $n = 6$ (total goblets),
• $k = 1$ (one "second"),
• $p = 0.1$ (probability of a goblet being a second),
• $1-p = 0.9$ (probability of a goblet not being a second).

$P(X = 1) = \binom{6}{1} \cdot (0.1)^1 \cdot (0.9)^5$

$P(X = 1) = 6 \cdot 0.1 \cdot (0.59049) = 0.354294$

Thus, the probability that only one is a second is approximately 0.354 (35.4%).

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Problem b: Probability that at least two are seconds

"At least two" means $P(X \geq 2)$. This can be calculated as:

$P(X \geq 2) = 1 - P(X < 2) = 1 - \big(P(X = 0) + P(X = 1)\big)$

We already know $P(X = 1) = 0.354294$. Now calculate $P(X = 0)$:

$P(X = 0) = \binom{6}{0} \cdot (0.1)^0 \cdot (0.9)^6 = 1 \cdot 1 \cdot 0.531441 = 0.531441$

So:

$P(X \geq 2) = 1 - (0.531441 + 0.354294) = 1 - 0.885735 = 0.114265$

The probability that at least two goblets are seconds is approximately 0.114 (11.4%).

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Problem c: Probability that at most five must be selected to find four non-seconds

This scenario can be modeled using the negative binomial distribution, which describes the probability of achieving $r$ successes (here, 4 non-seconds) after $k$ trials (here, selecting at most 5 goblets).

The probability is given by:

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

Here:

• $r = 4$ (non-seconds needed),
• $1-p = 0.9$ (probability of a non-second),
• $p = 0.1$ (probability of a second),
• $k$ ranges from $r$ to 5 (at most 5 selections).

We calculate for $k = 4$ and $k = 5$, then sum them.

Case 1: $k = 4$

$P(X = 4) = \binom{3}{3} \cdot (0.1)^0 \cdot (0.9)^4 = 1 \cdot 1 \cdot 0.6561 = 0.6561$

Case 2: $k = 5$

$P(X = 5) = \binom{4}{3} \cdot (0.1)^1 \cdot (0.9)^4 = 4 \cdot 0.1 \cdot 0.6561 = 0.26244$

Add the probabilities:

$P(\text{at most 5}) = P(X = 4) + P(X = 5) = 0.6561 + 0.26244 = 0.91854$

Thus, the probability is approximately 0.919 (91.9%).

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Summary of Results:

a. Probability that only one is a second: 0.354 (35.4%)
b. Probability that at least two are seconds: 0.114 (11.4%)
c. Probability that at most five must be selected to find four non-seconds: 0.919 (91.9%)

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Would you like further explanations on any of these parts?

Relative Questions:

1. How does the binomial formula work in practical situations?
2. What are the assumptions of the binomial distribution?
3. How does the negative binomial distribution differ from the binomial distribution?
4. Why do probabilities sum to 1 in these distributions?
5. How can these probability concepts be applied to quality control in manufacturing?

Tip:

Always verify that the total probabilities across all outcomes equal 1 as a sanity check for your calculations.