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

Cheyanne Leigh 2021-10-02 Answered
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)=_

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Expert Answer

irwchh
Answered 2021-10-03 Author has 10312 answers
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) \(\displaystyle={0.99}\)
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