Step 1

Given Data : Probability of no spelling mistakes on a page P (no mistake) \(\displaystyle={0.7}\)

To Find : Probability of at least one without mistakes

Let X be the number of pages among these 4 pages which Julia writes with zero mistakes

Binomial Probability : The probability of exactly x successes on n repeated trials in an experiment which has two possible outcomes. It express as :

Binomial Probability \(\displaystyle=^{{{n}}}{C}_{{{x}}}{p}^{{{x}}}{\left({1}-{p}\right)}^{{{n}-{x}}}\)

Step 2

In the given question we have :

Probability \(\displaystyle{p}={0.7}\)

n is the total number of pages i.e. \(\displaystyle{n}={4}\)

To find the at-least one page out of 4 pages with zero mistakes, \(\displaystyle{X}\geq{1}\)

Substitute the value of p,n and X in the expression of Binomial probability:

Step 3

Binomial Probability \(\displaystyle=^{{{n}}}{C}_{{{x}}}{p}^{{{x}}}{\left({1}-{p}\right)}^{{{n}-{x}}}\)

Probability for \(\displaystyle{\left({X}={x}\right)}=^{{{4}}}{C}_{{{x}}}{p}^{{{x}}}{\left({1}-{p}\right)}^{{{4}-{x}}}\)

for \(\displaystyle{x}\geq{1}\)

\(\displaystyle{P}{\left({x}\geq{1}\right)}={1}-{P}{\left({x}={0}\right)}\)

\(\displaystyle{P}{\left({x}\geq{1}\right)}={1}-^{{{4}}}{C}_{{{x}}}{\left({0.7}\right)}^{{{x}}}{\left({1}-{0.7}\right)}^{{{4}-{x}}}{p}{\left({x}\geq{1}\right)}={1}-^{{{4}}}{C}_{{{0}}}{\left({0.7}\right)}^{{{0}}}{\left({0.3}\right)}^{{{4}-{0}}}\)

\(\displaystyle{P}{\left({x}\geq{1}\right)}={1}-{\frac{{{4}!}}{{{0}!{\left({4}-{0}\right)}!}}}\times{1}\times{\left({0.3}\right)}^{{{4}}}\)

\(\displaystyle{P}{\left({x}\geq{1}\right)}={1}-{1}\times{1}\times{0.0081}\)

\(\displaystyle{P}{\left({x}\geq{1}\right)}={1}-{0.0081}\)

\(\displaystyle{P}{\left({x}\geq{1}\right)}={0.9919}\)

So, Probability of at least one page with zero mistake is 0.9919

Round of to the nearest hundredth 0.9919~0.99

Answer:

P(at-least one without mistake) \(\displaystyle={0.99}\)

Given Data : Probability of no spelling mistakes on a page P (no mistake) \(\displaystyle={0.7}\)

To Find : Probability of at least one without mistakes

Let X be the number of pages among these 4 pages which Julia writes with zero mistakes

Binomial Probability : The probability of exactly x successes on n repeated trials in an experiment which has two possible outcomes. It express as :

Binomial Probability \(\displaystyle=^{{{n}}}{C}_{{{x}}}{p}^{{{x}}}{\left({1}-{p}\right)}^{{{n}-{x}}}\)

Step 2

In the given question we have :

Probability \(\displaystyle{p}={0.7}\)

n is the total number of pages i.e. \(\displaystyle{n}={4}\)

To find the at-least one page out of 4 pages with zero mistakes, \(\displaystyle{X}\geq{1}\)

Substitute the value of p,n and X in the expression of Binomial probability:

Step 3

Binomial Probability \(\displaystyle=^{{{n}}}{C}_{{{x}}}{p}^{{{x}}}{\left({1}-{p}\right)}^{{{n}-{x}}}\)

Probability for \(\displaystyle{\left({X}={x}\right)}=^{{{4}}}{C}_{{{x}}}{p}^{{{x}}}{\left({1}-{p}\right)}^{{{4}-{x}}}\)

for \(\displaystyle{x}\geq{1}\)

\(\displaystyle{P}{\left({x}\geq{1}\right)}={1}-{P}{\left({x}={0}\right)}\)

\(\displaystyle{P}{\left({x}\geq{1}\right)}={1}-^{{{4}}}{C}_{{{x}}}{\left({0.7}\right)}^{{{x}}}{\left({1}-{0.7}\right)}^{{{4}-{x}}}{p}{\left({x}\geq{1}\right)}={1}-^{{{4}}}{C}_{{{0}}}{\left({0.7}\right)}^{{{0}}}{\left({0.3}\right)}^{{{4}-{0}}}\)

\(\displaystyle{P}{\left({x}\geq{1}\right)}={1}-{\frac{{{4}!}}{{{0}!{\left({4}-{0}\right)}!}}}\times{1}\times{\left({0.3}\right)}^{{{4}}}\)

\(\displaystyle{P}{\left({x}\geq{1}\right)}={1}-{1}\times{1}\times{0.0081}\)

\(\displaystyle{P}{\left({x}\geq{1}\right)}={1}-{0.0081}\)

\(\displaystyle{P}{\left({x}\geq{1}\right)}={0.9919}\)

So, Probability of at least one page with zero mistake is 0.9919

Round of to the nearest hundredth 0.9919~0.99

Answer:

P(at-least one without mistake) \(\displaystyle={0.99}\)