Maria Huey

2021-12-27

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!

### Answer & Explanation

Durst37

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, $P\left(A\right)=0.05$
Number of items, $n=75$
It is known that probability mass function of binomial distribution is,
$P\left(X=r\right){=}^{n}{C}_{r}\cdot {p}^{r}\cdot {\left(1-p\right)}^{n-r}$
Step 2
a) To find the probability of getting exactly 5 defective items.
Let's take in this case, $r=5$
$P\left(X=r\right){=}^{n}{C}_{r}\cdot {p}^{r}\cdot {\left(1-p\right)}^{n-r}$
$P\left(X=5\right){=}^{75}{C}_{5}\cdot {p}^{5}\cdot {\left(1-p\right)}^{75-5}$
$P\left(X=5\right)=\frac{75!}{\left(75-5\right)!5!}\cdot {\left(0.05\right)}^{5}\cdot {\left(1-0.05\right)}^{75-5}$
$P\left(X=5\right)=\frac{75!}{\left(70\right)!5!}\cdot {\left(0.05\right)}^{5}\cdot {\left(0.95\right)}^{70}$
$P\left(X=5\right)=\frac{75×74×73×72×71×70!}{\left(70\right)!5×3×2×1}\cdot 0.0000003125\cdot \left(0.027583\right)$
$P\left(X=5\right)=\frac{75×74×73×72×71}{5×3×2}\cdot 0.0000003125\cdot \left(0.027583\right)$
$P\left(X=5\right)=0.1488$
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, $r=0$.
For $r=0$, the probability mass function reduces to,
$P\left(X=r\right)={\left(1-p\right)}^{n-r}P\left(X=0\right)={\left(1-0.05\right)}^{75-0}$
$P\left(X=0\right)={\left(0.95\right)}^{75}$
$P\left(X=0\right)=0.0213$
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.
$P{\left(X=0\right)}^{c}=1-P\left(X=0\right)$
$P{\left(X=0\right)}^{c}=1-0.0213$
$P{\left(X=0\right)}^{c}=0.9787$
Hence the probability of getting atleast one defective item is 0.9787

Annie Levasseur

Step 1
The distribution of the number of defective items in this run is binomial with $n=75$ items and $p=0.05$ of each item being defective. The probability mass function of the binomial distribution is:
$P\left(X=k\right)=\left(nCk\right)×\left({p}^{k}\right)×{\left(1-p\right)}^{n-k}$
where nCk is the number of combinations of k objects chosen from
$n=\frac{n!}{k!×\left(n-k\right)!}$
1) In this case, we set and $p=0.05$, so:
$P\left(X=5\right)=\left(75C5\right)×\left({0.05}^{5}\right)×{\left(0.95\right)}^{70}$
$P\left(X=5\right)=\left(\frac{75!}{5!×\left(70\right)!}\right)×\left({0.05}^{5}\right)×{\left(0.95\right)}^{70}$
$P\left(X=5\right)=\left(\frac{75!}{5!×\left(70\right)!}\right)×\left({0.05}^{5}\right)×{\left(0.95\right)}^{70}=0.1488$
2) In this case, we set k = 0. The binomial probability mass function reduces to ${\left(1=p\right)}^{n}$ when $k=0$, so $P\left(X=0\right)={\left(0.95\right)}^{75}=0.0213$
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 $1-P\text{(original event)}$, so the probability of at least one defective item is $1-0.0213=0.9787$.

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,
The probability mass function of X is given below:
$P\left(X=x\right)=\left(nx\right){p}^{x}\left(1-p{\right)}^{n-x}$
a) Obtain the probability of getting exactly 5 defective items:
$P\left(X=5\right)=\left(755\right){0.05}^{5}\left(1-0.05{\right)}^{75-5}$
$=17,259,390×{0.05}^{5}\left(0.95{\right)}^{70}$
=0.1488
Thus, $P\left(\text{Exactly 5 defective items}\right)=0.1488$
Step 2
b) Obtain the probability of getting at least one defective item:
$P\left(X\ge 1\right)=1-P\left(X<1\right)$
$=1-P\left(X=0\right)$
$=1-\left(750\right){0.05}^{0}\left(1-0.05{\right)}^{75-0}$
$=1-\left(0.95{\right)}^{75}$
$=1-0.0213$
Thus, $P\left(\text{No defective items}\right)=0.0213.$
c) Obtain the probability of getting at least one defective item:
$P\left(X\ge 1\right)=1-P\left(X<1\right)$
$=1-P\left(X=0\right)$
$=1-\left(750\right){0.05}^{0}\left(1-0.05{\right)}^{75-0}$
$=1-\left(0.95{\right)}^{75}$
$=1-0.0213$
$=0.9787$
Thus, $P\left(\text{at least one defective item}\right)=0.9787$

Do you have a similar question?

Recalculate according to your conditions!