# When Julia is writing a first draft, there is 0.7 probability that there will be

When Julia is writing a first draft, there is 0.7 probability that there will be no spelling mistakes on a page. One day, Julia writes a first draft that is 4 pages long.
Assuming that Julia is equally likely to have a spelling mistake on each of the 4 pages, what is the probability that she will have no spelling mistakes on at least one of them?
P(at least one without mistakes)=_

• Questions are typically answered in as fast as 30 minutes

### Plainmath recommends

• Get a detailed answer even on the hardest topics.
• Ask an expert for a step-by-step guidance to learn to do it yourself.

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