Worksheet titled "Binomial and Geometric Worksheet" featuring six problems that require calculating probabilities using binomial and geometric distributions.

Final Answer:
0.0345

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Explanation:
We are told that, on average, 3% of items are defective. So the probability that any one item is defective is
\( p = 0.03 \), and the probability it is *not* defective is \( q = 1 - 0.03 = 0.97 \).

Each box has 12 items. The company wants no more than one defective item per box — that means the box fails the guarantee if it has 2 or more defective items.

So, the box *fails* if there are 2, 3, ..., up to 12 defective items.

It’s easier to find the complement:
Probability the box *satisfies* the guarantee =
P(0 defective) + P(1 defective)

Then:
P(box fails) = 1 − [P(0) + P(1)]

We use the binomial formula:
\[
P(k \text{ defectives}) = \binom{n}{k} p^k (1-p)^{n-k}
\]

Here, \( n = 12 \), \( p = 0.03 \)

- P(0) = \( \binom{12}{0} (0.03)^0 (0.97)^{12} = 1 \cdot 1 \cdot (0.97)^{12} \)
Calculate: \( 0.97^{12} \approx 0.6938 \)

- P(1) = \( \binom{12}{1} (0.03)^1 (0.97)^{11} = 12 \cdot 0.03 \cdot (0.97)^{11} \)
First, \( 0.97^{11} \approx 0.7153 \)
Then: \( 12 \cdot 0.03 = 0.36 \)
So P(1) ≈ \( 0.36 \cdot 0.7153 \approx 0.2575 \)

Add them:
P(satisfies) ≈ 0.6938 + 0.2575 = 0.9513

So P(fails) = 1 − 0.9513 = 0.0487

Wait — that doesn’t match the earlier answer. Let me double-check calculations more precisely.

Better to compute with more accuracy:

Use calculator-style:
- \( 0.97^{12} = (0.97)^{12} \approx 0.693842 \)
- \( 0.97^{11} = 0.693842 / 0.97 \approx 0.715301 \)
- P(1) = 12 × 0.03 × 0.715301 = 0.36 × 0.715301 = 0.257508

Sum = 0.693842 + 0.257508 = 0.95135

1 − 0.95135 = 0.04865

But the problem says “the box will fail to satisfy the guarantee” — i.e., more than 1 defective → 2 or more.

However, maybe the question expects using Poisson approximation? Let’s see:

For binomial with small p and moderate n, Poisson with λ = np = 12 × 0.03 = 0.36 is good.

Poisson:
P(k) = \( e^{-λ} \frac{λ^k}{k!} \)

P(0) = \( e^{-0.36} ≈ 0.6977 \)
P(1) = \( e^{-0.36} × 0.36 ≈ 0.6977 × 0.36 ≈ 0.2512 \)
Sum = 0.9489
Fail = 1 − 0.9489 = 0.0511

Still not 0.0345.

Wait — maybe I misread: “guarantee no more than one defective item per box” → box fails if it has more than one, i.e., ≥2. That’s what we did.

But let me recompute exact binomial with high precision:

Using exact computation:
- P(0) = (0.97)^12
Let me compute step by step:
0.97² = 0.9409
0.97⁴ = (0.9409)² = 0.88529281
0.97⁸ = (0.88529281)² ≈ 0.783743
0.97¹² = 0.97⁸ × 0.97⁴ ≈ 0.783743 × 0.88529281 ≈ 0.693842 (same)

P(1) = 12 × 0.03 × (0.97)^11
(0.97)^11 = (0.97)^12 / 0.97 = 0.693842 / 0.97 = 0.715301

12 × 0.03 = 0.36
0.36 × 0.715301 = 0.257508

Total success = 0.95135
Failure = 0.04865

Hmm. But the expected answer might be 0.0345 — could the problem mean “exactly one defective” is failure? No, wording says “no more than one” → 0 or 1 is OK.

Wait — maybe the process produces *on average* 3% defective, but they ship 12 items, and want guarantee of *at most one defective*, and ask: probability box fails → i.e., >1 defective.

Let me check if perhaps they used p = 0.03, n = 12, and computed P(X ≥ 2) directly via software:

Using binomial CDF:
P(X ≤ 1) = binomcdf(12, 0.03, 1) = ?

I can verify with known values:
In many textbooks, for n=12, p=0.03:
P(0) = 0.6938
P(1) = 0.2575
P(2) = C(12,2)*(0.03)^2*(0.97)^10
C(12,2)=66
(0.03)^2 = 0.0009
(0.97)^10 = (0.97)^12 / (0.97)^2 = 0.693842 / 0.9409 ≈ 0.7374
So P(2) ≈ 66 × 0.0009 × 0.7374 = 0.0594 × 0.7374? Wait: 66×0.0009 = 0.0594
0.0594 × 0.7374 ≈ 0.0438

Then P(≥2) = 1 − (0.6938+0.2575) = 0.0487, and P(2) alone is ~0.0438, so rest (3+) is ~0.0049 — makes sense.

So correct answer is approximately 0.0487, not 0.0345.

But the instruction says: “Solve the problem accurately.” And user gave no answer key.

Wait — maybe I misread the number of items? “ships 12 items in each box” — yes.

Unless… the 3% is *per batch*, not per item? No, standard interpretation is per item.

Let me consider: Could the question be asking for probability that the box fails — meaning it has more than one, but maybe they consider “fail” as having at least one? That would be 1 − P(0) = 1 − 0.6938 = 0.3062 — no.

Alternatively, maybe the guarantee is “no defective items”, and “no more than one” is misread? But wording is clear: “guarantee no more than one defective item per box”.

Given all this, the accurate answer is ≈0.0487, which rounds to 0.049 or 0.0487.

But the initial “Final Answer” I wrote was 0.0345 — that was a mistake.

Since the instruction says *accurately*, I must correct it.

Let me recalculate with a calculator (simulate):

Using Python-like:
from math import comb
p = 0.03
n = 12
p0 = (1-p)**n
p1 = n * p * (1-p)**(n-1)
p_fail = 1 - (p0 + p1)
p0 = 0.97**12 = 0.693842
p1 = 12 * 0.03 * 0.97**11 = 12*0.03*0.715301 = 0.257508
sum = 0.95135
p_fail = 0.04865

So final answer: 0.0487 (to 4 decimal places)

But the problem may expect 4 decimal places.

I will give the accurate value:

Final Answer:
0.0487