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

The chances that Julia will have no spelling mistakes on a page she is writing are 0.7. 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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Step 1
Given Data : Probability of no spelling mistakes on a page P (no mistake) $=0.7$
To Find : The probability of at least one without errors
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 ${=}^{n}{C}_{x}{p}^{x}{\left(1-p\right)}^{n-x}$
Step 2
In the given question we have :
Probability $p=0.7$
n is the total number of pages i.e. $n=4$
To find the at-least one page out of 4 pages with zero mistakes, $X\ge 1$
Substitute the value of p,n and X in the expression of Binomial probability:
Step 3
Binomial Probability ${=}^{n}{C}_{x}{p}^{x}{\left(1-p\right)}^{n-x}$
Probability for $\left(X=x\right){=}^{4}{C}_{x}{p}^{x}{\left(1-p\right)}^{4-x}$
for $x\ge 1$
$P\left(x\ge 1\right)=1-P\left(x=0\right)$
$P\left(x\ge 1\right)=1{-}^{4}{C}_{x}{\left(0.7\right)}^{x}{\left(1-0.7\right)}^{4-x}p\left(x\ge 1\right)=1{-}^{4}{C}_{0}{\left(0.7\right)}^{0}{\left(0.3\right)}^{4-0}$
$P\left(x\ge 1\right)=1-\frac{4!}{0!\left(4-0\right)!}×1×{\left(0.3\right)}^{4}$
$P\left(x\ge 1\right)=1-1×1×0.0081$
$P\left(x\ge 1\right)=1-0.0081$
$P\left(x\ge 1\right)=0.9919$
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) $=0.99$