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