Probability: breaking keyboard

Question

I'm trying to self-learn theory of probability, I came across the following basic problem that I think I solved but I'm not sure as I'm very new to this.

Problem: A keyboard manufacturer states that his keyboards may miss a key press in one of 1000 key presses.

If a key is pressed 8000 times, what is the probability of

1. Exactly two key presses are missed
2. One or more key presses are missed

My solution:

Let $A$ = { a keypress is missed in 1000 attempts }

Let $B$ = { A happened exactly two times in 8 attempts }

$P(A)$ = $\frac 1{1000}$

$P(\overline A) = (1 - \frac 1{1000})$

$P(B)$ = ?

If the sequence $A_1$, $A_2$, $A_3$, $A_4$, $A_5$, $A_6$, $A_7$, $A_8$ are the 8 attempts of $A$

then let's choose a combination, in which A happened two times and it didn't happen 6 times from the sequence above

e.g $A$$A$$\overline A$$\overline A$$\overline A$$\overline A$$\overline A$$\overline A$

In that particular combination

$P(AA\overline A\overline A\overline A\overline A\overline A\overline A) = P(A)^2P(\overline A)^6$

For every other combination from the sequence in which A happens twice and it didn't happen 6 times the probability is the same (is this how I'm supposed to write it? Do I have to state that the combinations are unrelated to themselves?)

Since there are $\binom 8 2$ combinations that apply then we can conclude that

$P(B) = \binom 8 2 P(A)^2P(\overline A)^6 = \binom 8 2 \frac 1{1000}^2(1 - \frac1 {1000})^6$

For the second part:

Let $C$ = { $A$ never happened in 8 attempts }

// opposite of "A happened at least once is 8 attempts"

Then what is being searched for is $1 - P(C)$

$P(C) = P(\overline A)^8 = (1 - \frac1 {1000})^8$

$1 - P(C) = 1 - (1 - \frac1 {1000})^8$

Is this correct? Is it the proper way to write the solution? If not, could somebody write a proper way of solving this solution even if mine is correct.

Answer 1

A side note. We agree that $P(X\geq 1)=1-\left(1-\frac{1}{1000} \right)^{8000}$

This term can be approximated. For large n

$\left(1-\frac{x}{n} \right)^{n}\approx e^{-x}$

$\texttt{The fraction can be expanded by 8}$

$\left(1-\frac{8}{8000} \right)^{8000}\approx e^{-8}$

Thus $P(X\geq 1)\approx 1-e^{-8}=99.97\%$

Answer 2

A keyboard manufacturer states that his keyboards may miss a key press in one of 1000 key presses.

That means that the probability on missing a key press is $0.001$.

You are dealing with binomial distribution with parameteres $n=8000$ and $p=0.001$.

If $X$ denotes the number of missed key presses then to be found are:

• $P(X=2)=\binom{n}2p^2(1-p)^{n-2}$
• $P(X\geq1)=1-P(X=0)=1-\binom{n}0p^0(1-p)^{n}=1-(1-p)^{n}$