Question

Contaminated gun cartridges. A weapons manufacturer uses liquid fuel to produce gun cartridges. The fuel can get mixed with another liquid to produce a contaminated cartridge during the manufacturing process. A University of South Florida statistician hired by the company to investigate the level of contamination in the stored cartridges found that 23% of the cartridges in a particular lot were contaminated. Suppose you randomly sample (without replacement) gun cartridges from this lot until you find a contaminated one. Let x be the number of cartridges sampled until a contaminated one is found. It is known that the formula gives the probability distribution for x

$\mathbf{p}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\left(}\mathbf{.23}\mathbf{\right)}{\mathbf{\left(}\mathbf{.77}\mathbf{\right)}}^{\mathbf{x}\mathbf{-1}}\mathbf{,}\mathbf{x}\mathbf{=1,2,3}$

1. Find $\mathbf{\text{p}}\mathbf{\left(}\mathbf{\text{1}}\mathbf{\right)}$. Interpret this result.
2. Find $\mathbf{\text{p}}\mathbf{\left(}\mathbf{\text{5}}\mathbf{\right)}$. Interpret this result.
3. Find $\mathbf{\text{p}}\mathbf{(}\mathbf{\text{x}}\mathbf{\ge }\mathbf{\text{2}}\mathbf{)}$. Interpret this result.

Short answer

1. 0.23
2. 0.08
3. 0.41

Step-by-step solution

Calculation of  p(1)

The probability of finding 1 contaminated gun cartridge is calculated below:

$$\begin{gathered} \mathbf{\text{p}}\mathbf{\left(}\mathbf{\text{x}}\mathbf{\right)}\mathbf{\text{=}}\mathbf{\left(}\mathbf{\text{.23}}\mathbf{\right)}{\mathbf{\left(}\mathbf{\text{.77}}\mathbf{\right)}}^{\mathbf{\text{x}}\mathbf{-}\mathbf{\text{1}}} \\ \mathbf{\text{p}}\mathbf{\left(}\mathbf{\text{1}}\mathbf{\right)}\mathbf{\text{=}}\mathbf{\left(}\mathbf{\text{.23}}\mathbf{\right)}{\mathbf{\left(}\mathbf{\text{.77}}\mathbf{\right)}}^{\mathbf{-}\mathbf{\text{1}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=}}\mathbf{\left(}\mathbf{\text{.23}}\mathbf{\right)}{\mathbf{\left(}\mathbf{\text{.77}}\mathbf{\right)}}^{\mathbf{\text{0}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=0.23}} \end{gathered}$$

Interpretation of the value of p(1)

Here the value $\mathbf{\text{p}}\mathbf{\left(}\mathbf{\text{1}}\mathbf{\right)}$from the calculation is 0.23.This value indicates that the probability of finding a contaminated cartridge from the sample of cartridges is 0.23.

(b) Calculation of p(5)

The probability of finding 5 contaminated cartridges from the sample is calculated below:

$$\begin{gathered} \mathbf{\text{p}}\mathbf{\left(}\mathbf{\text{x}}\mathbf{\right)}\mathbf{\text{=}}\mathbf{\left(}\mathbf{\text{.23}}\mathbf{\right)}{\mathbf{\left(}\mathbf{\text{.77}}\mathbf{\right)}}^{\mathbf{\text{x-1}}} \\ \mathbf{\text{p}}\mathbf{\left(}\mathbf{\text{5}}\mathbf{\right)}\mathbf{\text{=}}\mathbf{\left(}\mathbf{\text{.23}}\mathbf{\right)}{\mathbf{\left(}\mathbf{\text{.77}}\mathbf{\right)}}^{\mathbf{\text{5-1}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=}}\mathbf{\left(}\mathbf{\text{.23}}\mathbf{\right)}{\mathbf{\left(}\mathbf{\text{.77}}\mathbf{\right)}}^{\mathbf{\text{4}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=0.08}} \end{gathered}$$

Interpretation of the value of p(5)

Here the value of $\mathbf{p}\mathbf{\left(}\mathbf{5}\mathbf{\right)}$ the above calculation is found to be 0.08.This value indicates that the probability of finding five contaminated gun cartridges from the sample of several cartridges is 0.08.

(c) Calculation of p(x≥2)

The probability of finding a maximum of two contaminated cartridges is calculated below:

$$\begin{gathered} \mathbf{\text{p}}\mathbf{(}\mathbf{\text{x}}\mathbf{\ge }\mathbf{\text{2}}\mathbf{)}\mathbf{\text{=}}\mathbf{\left(}\mathbf{\text{.23}}\mathbf{\right)}{\mathbf{\left(}\mathbf{\text{.77}}\mathbf{\right)}}^{\mathbf{\text{x-1}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{p}}\mathbf{\left(}\mathbf{\text{1}}\mathbf{\right)}\mathbf{\text{=}}\mathbf{\left(}\mathbf{\text{.23}}\mathbf{\right)}{\mathbf{\left(}\mathbf{\text{.77}}\mathbf{\right)}}^{\mathbf{\text{1-1}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=}}\mathbf{\left(}\mathbf{\text{.23}}\mathbf{\right)}{\mathbf{\left(}\mathbf{\text{.77}}\mathbf{\right)}}^{\mathbf{\text{0}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=0.23}} \end{gathered}$$

$$\begin{gathered} \mathbf{\text{p}}\mathbf{\left(}\mathbf{\text{2}}\mathbf{\right)}\mathbf{\text{=}}\mathbf{\left(}\mathbf{\text{.23}}\mathbf{\right)}{\mathbf{\left(}\mathbf{\text{.77}}\mathbf{\right)}}^{\mathbf{\text{2-1}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=}}\mathbf{\left(}\mathbf{\text{.23}}\mathbf{\right)}{\mathbf{\left(}\mathbf{\text{.77}}\mathbf{\right)}}^{\mathbf{\text{1}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=0.18}} \\ \mathbf{\text{p}}\mathbf{(}\mathbf{\text{x}}\mathbf{\ge }\mathbf{\text{2}}\mathbf{)}\mathbf{\text{=p}}\mathbf{\left(}\mathbf{\text{1}}\mathbf{\right)}\mathbf{\text{+p}}\mathbf{\left(}\mathbf{\text{2}}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=0.23+0.18}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=0.41}} \end{gathered}$$

Interpretation of the value of p(x≥2)

Here, the calculation$\mathbf{\text{p}}\mathbf{(}\mathbf{\text{x}}\mathbf{\ge }\mathbf{\text{2}}\mathbf{)}$is found to be 0.41.This value indicates that the probability of finding a maximum of two contaminated gun cartridges by the researchers from the sample is 0.41.