The probability that a certain machine turns out a defective

Maria Huey 2021-12-27 Answered
The probability that a certain machine turns out a defective item is 5%. Find the probabilities that in a set of 75 items:
(a) Exactly 5 defective items
(b) No defective items
(c) At least one defective item
(d) What is the expected value of the number of defective items?
CANNOT BE EXCEL! Thank you!

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Expert Answer

Durst37
Answered 2021-12-28 Author has 3385 answers
Step 1
As per bartleby guidelines only first three subparts are to be solved. Please upload other parts separately.
Given:
Probability of defective item is, \(\displaystyle{P}{\left({A}\right)}={0.05}\)
Number of items, \(\displaystyle{n}={75}\)
It is known that probability mass function of binomial distribution is,
\(\displaystyle{P}{\left({X}={r}\right)}=^{{{n}}}{C}_{{{r}}}\cdot{p}^{{{r}}}\cdot{\left({1}-{p}\right)}^{{{n}-{r}}}\)
Step 2
a) To find the probability of getting exactly 5 defective items.
Let's take in this case, \(\displaystyle{r}={5}\)
\(\displaystyle{P}{\left({X}={r}\right)}=^{{{n}}}{C}_{{{r}}}\cdot{p}^{{{r}}}\cdot{\left({1}-{p}\right)}^{{{n}-{r}}}\)
\(\displaystyle{P}{\left({X}={5}\right)}=^{{{75}}}{C}_{{{5}}}\cdot{p}^{{{5}}}\cdot{\left({1}-{p}\right)}^{{{75}-{5}}}\)
\(\displaystyle{P}{\left({X}={5}\right)}={\frac{{{75}!}}{{{\left({75}-{5}\right)}!{5}!}}}\cdot{\left({0.05}\right)}^{{{5}}}\cdot{\left({1}-{0.05}\right)}^{{{75}-{5}}}\)
\(\displaystyle{P}{\left({X}={5}\right)}={\frac{{{75}!}}{{{\left({70}\right)}!{5}!}}}\cdot{\left({0.05}\right)}^{{{5}}}\cdot{\left({0.95}\right)}^{{{70}}}\)
\(\displaystyle{P}{\left({X}={5}\right)}={\frac{{{75}\times{74}\times{73}\times{72}\times{71}\times{70}!}}{{{\left({70}\right)}!{5}\times{3}\times{2}\times{1}}}}\cdot{0.0000003125}\cdot{\left({0.027583}\right)}\)
\(\displaystyle{P}{\left({X}={5}\right)}={\frac{{{75}\times{74}\times{73}\times{72}\times{71}}}{{{5}\times{3}\times{2}}}}\cdot{0.0000003125}\cdot{\left({0.027583}\right)}\)
\(\displaystyle{P}{\left({X}={5}\right)}={0.1488}\)
Hence, the probability of getting exactly 5 defective items is 0.1488
Step 3
b) To find the probability of getting no defective items.
Let's take in this case, \(\displaystyle{r}={0}\).
For \(\displaystyle{r}={0}\), the probability mass function reduces to,
\(\displaystyle{P}{\left({X}={r}\right)}={\left({1}-{p}\right)}^{{{n}-{r}}}{P}{\left({X}={0}\right)}={\left({1}-{0.05}\right)}^{{{75}-{0}}}\)
\(\displaystyle{P}{\left({X}={0}\right)}={\left({0.95}\right)}^{{{75}}}\)
\(\displaystyle{P}{\left({X}={0}\right)}={0.0213}\)
Hence, the probability of getting no defective items is 0.0213
Step 4
c) The probability of getting atleast one defective item is the complement of probability of no defective item.
\(\displaystyle{P}{\left({X}={0}\right)}^{{{c}}}={1}-{P}{\left({X}={0}\right)}\)
\(\displaystyle{P}{\left({X}={0}\right)}^{{{c}}}={1}-{0.0213}\)
\(\displaystyle{P}{\left({X}={0}\right)}^{{{c}}}={0.9787}\)
Hence the probability of getting atleast one defective item is 0.9787
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Annie Levasseur
Answered 2021-12-29 Author has 2743 answers
Step 1
The distribution of the number of defective items in this run is binomial with \(\displaystyle{n}={75}\) items and \(\displaystyle{p}={0.05}\) of each item being defective. The probability mass function of the binomial distribution is:
\(\displaystyle{P}{\left({X}={k}\right)}={\left({n}{C}{k}\right)}\times{\left({p}^{{{k}}}\right)}\times{\left({1}-{p}\right)}^{{{n}-{k}}}\)
where nCk is the number of combinations of k objects chosen from
\(\displaystyle{n}={\frac{{{n}!}}{{{k}!\times{\left({n}-{k}\right)}!}}}\)
1) In this case, we set \(\displaystyle{k}={5},\ {n}={75}\) and \(\displaystyle{p}={0.05}\), so:
\(\displaystyle{P}{\left({X}={5}\right)}={\left({75}{C}{5}\right)}\times{\left({0.05}^{{{5}}}\right)}\times{\left({0.95}\right)}^{{{70}}}\)
\(\displaystyle{P}{\left({X}={5}\right)}={\left({\frac{{{75}!}}{{{5}!\times{\left({70}\right)}!}}}\right)}\times{\left({0.05}^{{{5}}}\right)}\times{\left({0.95}\right)}^{{{70}}}\)
\(\displaystyle{P}{\left({X}={5}\right)}={\left({\frac{{{75}!}}{{{5}!\times{\left({70}\right)}!}}}\right)}\times{\left({0.05}^{{{5}}}\right)}\times{\left({0.95}\right)}^{{{70}}}={0.1488}\)
2) In this case, we set k = 0. The binomial probability mass function reduces to \(\displaystyle{\left({1}={p}\right)}^{{{n}}}\) when \(\displaystyle{k}={0}\), so \(\displaystyle{P}{\left({X}={0}\right)}={\left({0.95}\right)}^{{{75}}}={0.0213}\)
3) The event that at least one item is defective is the compliment of the event that there are no defective items. The probability of a complimentary event happening is \(\displaystyle{1}-{P}\text{(original event)}\), so the probability of at least one defective item is \(\displaystyle{1}-{0.0213}={0.9787}\).
0
karton
Answered 2022-01-04 Author has 8659 answers

Step 1
Let X denotes the number of defective items in a machine which follows binomial distribution with the probability of success 0.05 the number of items selected is 75. That is, \(X\sim P(75,\ 0.05)\)
The probability mass function of X is given below:
\(P(X=x)=(nx)p^{x}(1-p)^{n-x}\)
a) Obtain the probability of getting exactly 5 defective items:
\(P(X=5)=(755)0.05^{5}(1-0.05)^{75-5}\)
\(=17,259,390\times0.05^{5}(0.95)^{70}\)
=0.1488
Thus, \(P(\text{Exactly 5 defective items})=0.1488\)
Step 2
b) Obtain the probability of getting at least one defective item:
\(P(X\geq1)=1-P(X<1)\)
\(=1-P(X=0)\)
\(=1-(750)0.05^{0}(1-0.05)^{75-0}\)
\(=1-(0.95)^{75}\)
\(=1-0.0213\)
Thus, \(P(\text{No defective items})=0.0213. \)
c) Obtain the probability of getting at least one defective item:
\(P(X\geq1)=1-P(X<1)\)
\(=1-P(X=0)\)
\(=1-(750)0.05^{0}(1-0.05)^{75-0}\)
\(=1-(0.95)^{75}\)
\(=1-0.0213\)
\(=0.9787\)
Thus, \(P(\text{at least one defective item})=0.9787\)

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