Given Information is: There are total of $10$ computers out of them $4$ are functional and rest of them are non-functional. Ofiicials are going to inspect $4$ computers that are chossen at random without replacement.
The officials are going to inspect 4 computers, think of its like $4$ trails. Success in each trail in means the computer we got for the trail is functional and here the most important point is success of these trials are not independent of the success in the other trails.
$\begin{align}P(\text{success in 2nd trail}) &= P(\text{success in 1st trail}) P(\text{success in 2nd trail | success in 1st trial}) + P(\text{failure in 1st trail})P(\text{success in 2nd trail | failure in first trials}) \\ &= \frac{4}{10}\times\frac{3}{9} + \frac{6}{10}\times\frac{4}{9}\\ &= \frac{2}{5}\end{align}$
$P(\text{success in 2nd trail | success in 1st trial}) = \dfrac{3}{9} = \dfrac{1}{3}$.
So the success in each trail is not independent of other success in other trails.
This implies, that we can't use binomial distribution there, because in binomial distribution the trails must be $\textbf{independent and identically distributed Bern(p) trails}$. But here are trails are not even independent, its cause we are doing without replacement, if there single word "without replacement" is removed then we are all set to use binomial distribution, but sadly we can't do that.
Define a random variable $X = \text{number of success in the 4 trials.}$
Note:-
- Here picking any 4 computers of inspection is equally likely, which allows us to use naive definition of probability.
System is deemed functional $\iff P(X\geq 3 ) = P(X = 3) + P(X = 4)$
$\begin{align}P(X \geq 3) &= P(X = 3) + P(X = 4)\\ &= \dfrac{\displaystyle\binom{4}{3}\binom{6}{1}}{\displaystyle\binom{10}{4}} + \dfrac{\displaystyle\binom{4}{4}}{\displaystyle\binom{10}{4}} \\ &\approx 0.119047\end{align}$
The answer is, $100\times 0.119047 = 11.9047$
Additional Information:
$\textbf{Bernouli Trail:-}$ An experiment that can result in either $\textbf{success}$ or $\textbf{failure}$ but not both, is called bernouli trial.
$\textbf{Story of Binomial: Bin(n,p)}$
I have a coin and i am going to toss it for $n$ times. $P(\text{Heads}) = p \text{ and } P(tails) = 1-p = q$.
Consider each toss as a trial, I am defining my $\textbf{success}$ in each trial as getting $\text{"Heads"}$ in that trail.
Let $X = \text{number of success in n trails}$
Here the trails are independent in the sense the outcomes of those trails is not going to effect our beliefs our outcomes of rest the of the trails and most important all these trails are bernouli trails. More precisely, all these trails are $\textbf{n independent Bern(p) trails.}$ Then we can say $X \sim \textbf{Bin(n,p)}$
So whenever you come across a situation where, all the trails are $\textbf{Independent}$ and all these trails follow $\textbf{Bern(p)}$ distribution then we can use $\textbf{Bin(n,p)}$.
- All the $n$ trails must be $\textbf{Independent}$
- Every trails must follow same $\textbf{Bern(p)}$ distribution.
Then only we can use $\textbf{Binomial(n,p)}$.
$\textbf{Story of HyperGeomtric: HGeo(w,b,n)}$
Imagine a box contains $w$ white balls and $b$ black balls. Now we are going to grab a sample of $n$ balls from the box without replacement. Now i am intersted in how many white balls in the sampled ones.
Now Consider each grab as a trail, i am defining my $\textbf{success}$ in each grab as picking a $\textbf{white}$ ball.
Let $X = \text{number of success in each trail.}$
Note:
- Here the trails are not independent (think why?)
- Even though each trail is bernouli trail, but each trial don't have same $\textbf{Bern(p)}$ distribution.
So its obvious that $X$ doen't binomial distribution. If you observe carefully the structure of this story and structure of the given problem is same, consider $w$ as functional and $b$ as non-functional.
Here $P(X = k ) = \dfrac{\displaystyle\binom{w}{k} \displaystyle\binom{b}{n-k}}{\displaystyle\binom{w+b}{n}}$ if $0\leq k\leq w \text{and} 0\leq n-k\leq b$ otherwise $P(X = k) = 0$.
