The probability that a certain machine turns out a defective item is 5%. Find the...

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,
${\displaystyle P(X=r){=}^n{C}_r\cdot p^r\cdot {(1-p)}^{n-r}}$
Step 2
a) To find the probability of getting exactly 5 defective items.
Let's take in this case, ${\displaystyle r=5}$
${\displaystyle P(X=r){=}^n{C}_r\cdot p^r\cdot {(1-p)}^{n-r}}$
${\displaystyle P(X=5){=}^{75}{C}_5\cdot p^5\cdot {(1-p)}^{75-5}}$
${\displaystyle P(X=5)=\frac{75!}{(75-5)!5!}\cdot {\left(0.05\right)}^5\cdot {(1-0.05)}^{75-5}}$
${\displaystyle P(X=5)=\frac{75!}{\left(70\right)!5!}\cdot {\left(0.05\right)}^5\cdot {\left(0.95\right)}^{70}}$
${\displaystyle P(X=5)=\frac{75\times 74\times 73\times 72\times 71\times 70!}{\left(70\right)!5\times 3\times 2\times 1}\cdot 0.0000003125\cdot \left(0.027583\right)}$
${\displaystyle P(X=5)=\frac{75\times 74\times 73\times 72\times 71}{5\times 3\times 2}\cdot 0.0000003125\cdot \left(0.027583\right)}$
${\displaystyle P(X=5)=0.1488}$
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,
${\displaystyle P(X=r)={(1-p)}^{n-r}P(X=0)={(1-0.05)}^{75-0}}$
${\displaystyle P(X=0)={\left(0.95\right)}^{75}}$
${\displaystyle P(X=0)=0.0213}$
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.
${\displaystyle P{(X=0)}^c=1-P(X=0)}$
${\displaystyle P{(X=0)}^c=1-0.0213}$
${\displaystyle P{(X=0)}^c=0.9787}$
Hence the probability of getting atleast one defective item is 0.9787

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:
${\displaystyle P(X=k)=\left(nCk\right)\times \left(p^k\right)\times {(1-p)}^{n-k}}$
where nCk is the number of combinations of k objects chosen from
${\displaystyle n=\frac{n!}{k!\times (n-k)!}}$
1) In this case, we set ${\displaystyle k=5,n=75}$ and ${\displaystyle p=0.05}$, so:
${\displaystyle P(X=5)=\left(75C5\right)\times \left({0.05}^5\right)\times {\left(0.95\right)}^{70}}$
${\displaystyle P(X=5)=\left(\frac{75!}{5!\times \left(70\right)!}\right)\times \left({0.05}^5\right)\times {\left(0.95\right)}^{70}}$
${\displaystyle P(X=5)=\left(\frac{75!}{5!\times \left(70\right)!}\right)\times \left({0.05}^5\right)\times {\left(0.95\right)}^{70}=0.1488}$
2) In this case, we set k = 0. The binomial probability mass function reduces to ${\displaystyle {(1=p)}^n}$ when ${\displaystyle k=0}$, so ${\displaystyle P(X=0)={\left(0.95\right)}^{75}=0.0213}$
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 ${\displaystyle 1-0.0213=0.9787}$.

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:
$P(X=5)=(755){0.05}^5(1-0.05{)}^{75-5}$
$=17,259,390\times {0.05}^5(0.95{)}^{70}$
=0.1488
Thus, $P(\text{Exactly 5 defective items})=0.1488$
Step 2
b) Obtain the probability of getting at least one defective item:
$P(X\ge 1)=1-P(X<1)$
$=1-P(X=0)$
$=1-(750){0.05}^0(1-0.05{)}^{75-0}$
$=1-(0.95{)}^{75}$
$=1-0.0213$
Thus, $P(\text{No defective items})=0.0213.$
c) Obtain the probability of getting at least one defective item:
$P(X\ge 1)=1-P(X<1)$
$=1-P(X=0)$
$=1-(750){0.05}^0(1-0.05{)}^{75-0}$
$=1-(0.95{)}^{75}$
$=1-0.0213$
$=0.9787$
Thus, $P(\text{at least one defective item})=0.9787$