 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?
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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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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 $$\displaystyle{1}-{0.0213}={0.9787}$$. karton

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, $$P(\text{at least one defective item})=0.9787$$