Question

find the indicated probabilities using the geometric distribution, the poisson distribution, or the binomial distribution. then determine if the events are unusual. if convenient, use the appropriate probability table or technology to find the probabilities.
a glass manufacturer finds that 1 in every 1000 glass items produced is warped. find the probability that (a) the first warped glass item is the 11th item produced, (b) the first warped item is the first, second, or third item produced, and (c) none of the first 10 glass items produced are defective.
(a) p(the first warped glass item is the 11th item produced) =
(round to three decimal places as needed.)

Explanation:

Step1: Identify the probability of success (p)

The probability that a glass item is warped is \( p = \frac{1}{1000} = 0.001 \). The probability of failure (not warped) is \( q = 1 - p = 1 - 0.001 = 0.999 \).

Step2: Apply the geometric distribution formula

The geometric distribution formula for the probability that the first success occurs on the \( k \)-th trial is \( P(X = k) = q^{k - 1} \cdot p \). Here, \( k = 11 \), so we substitute \( k = 11 \), \( q = 0.999 \), and \( p = 0.001 \) into the formula.

\[ \begin{align*} P(X = 11) &= (0.999)^{11 - 1} \cdot 0.001 \\ &= (0.999)^{10} \cdot 0.001 \end{align*} \]

Step3: Calculate \( (0.999)^{10} \)

Using a calculator, \( (0.999)^{10} \approx 0.990045 \).

Step4: Multiply by \( p \)

\[ \begin{align*} P(X = 11) &\approx 0.990045 \cdot 0.001 \\ &\approx 0.000990045 \end{align*} \]

Step5: Round to three decimal places

Rounding \( 0.000990045 \) to three decimal places gives \( 0.001 \). Wait, that seems off. Wait, let's recalculate \( (0.999)^{10} \) more accurately.

Using a calculator: \( 0.999^{10} = e^{10 \ln(0.999)} \approx e^{10 \times (-0.0010005)} \approx e^{-0.010005} \approx 0.9900498 \). Then multiply by 0.001: \( 0.9900498 \times 0.001 = 0.0009900498 \), which rounds to \( 0.001 \) when rounded to three decimal places? Wait, no, 0.0009900498 is approximately 0.001 when rounded to three decimal places? Wait, 0.0009900498 is 0.000990..., so to three decimal places, it's 0.001? Wait, no, the third decimal place is 0, the fourth is 9, so we round up the third decimal place? Wait, 0.0009900498: the first decimal place is 0, second is 0, third is 0, fourth is 9. Wait, no, 0.0009900498 is 0.000 (thousandths place) with the next digit 9, so when rounding to three decimal places, we look at the fourth decimal place (9), so we round the third decimal place (0) up by 1? Wait, no, 0.0009900498 is 0.000990..., so in decimal form, it's 0.0009900498. So the first decimal place: 0 (tenths), second: 0 (hundredths), third: 0 (thousandths), fourth: 9 (ten - thousandths). So when rounding to three decimal places, we look at the fourth decimal place (9) to round the third decimal place (0). So 0.000 + 0.001 (because 9 >= 5) = 0.001? Wait, but 0.0009900498 is approximately 0.001 when rounded to three decimal places? Wait, maybe my initial calculation of \( (0.999)^{10} \) is wrong. Wait, let's calculate \( 0.999^{10} \) step by step:

\( 0.999^2 = 0.998001 \)

\( 0.999^4 = (0.998001)^2 \approx 0.996005996 \)

\( 0.999^8 = (0.996005996)^2 \approx 0.992031888 \)

\( 0.999^{10} = 0.999^8 \times 0.999^2 \approx 0.992031888 \times 0.998001 \approx 0.9900459 \)

Then \( 0.9900459 \times 0.001 = 0.0009900459 \), which is approximately 0.001 when rounded to three decimal places? Wait, no, 0.0009900459 is 0.000990..., so the third decimal place is 0, the fourth is 9, so we round the third decimal place up: 0.001. Wait, but that seems counterintuitive. Wait, maybe the problem is that the probability of success is 0.001, so the probability that the first success is on the 11th trial is very small, but when rounded to three decimal places, it's 0.001? Wait, no, 0.0009900459 is approximately 0.001 when rounded to three decimal places? Let me check with a calculator. If I calculate \( (0.999)^{10} \times 0.001 \):

\( 0.999^{10} = e^{10 \ln(0.999)} \approx e^{10 \times (-0.0010005003)} \approx e^{-0.010005003} \approx 0.9900498 \)

Then \( 0.9900498 \times 0.001 = 0.0009900498 \), which is 0.0009900498. Rounding to three decimal places: look at the fourth decimal place, which is 9. So we round the third decimal place (0) up by 1, so 0.001. Wait, but 0.0009900498 is 0.000990..., so the third decimal place is 0, the fourth is 9, so yes, we round up the third decimal place to 1, making it 0.001.

Answer:

\( 0.001 \)