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!

Durst37

Expert

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

Expert

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

Expert

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

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