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A lot of 100 watches are known to have 10 defective watches. If 8 watches are selected (one by one with replacement) at random, the probability that there are at least one defective watch is ${\text{ = 1 - }}{\left( {\dfrac{9}{{10}}} \right)^x}$, then the value of x is ………………………………….

Last updated date: 19th Jul 2024
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Hint – Use probability distribution of random variables. Probability distribution provides the possibility of presence of different outputs.

In a lot of 100 watches we know that 10 are defective.
We have to select 8 watches one by one without replacement.
Let X denote the number of defective watches in 8 draws and let P be the probability of selecting a defective watch in a draw.
Now, X follows binomial distribution with parameters ${\text{n = 8}}$ and ${\text{p = }}\dfrac{{10}}{{100}} = \dfrac{1}{{10}}$ because we have total 100 watches out of which 10 are defective.
Now ${\text{P}}\left( {X = r} \right) = {}^n{c_r}{\left( p \right)^r}{\left( {1 - p} \right)^{n - r}}$
Using the above concept
$P\left( {X = r} \right) = {}^8{c_r}{\left( {\dfrac{1}{{10}}} \right)^r}{\left( {\dfrac{9}{{10}}} \right)^{8 - r}}$Where our ${\text{where our r = 0,1,2}}....{\text{8}}$
Now we are asked to find the probability that at least one defective watch is drawn.
So we have to find ${\text{P}}\left( {X > = 1} \right)$
Now ${\text{P}}\left( {X > = 1} \right) = 1 - P\left( {X = 0} \right)$
This is equal to ${\text{1 - }}{}^8{c_0}{\left( {\dfrac{1}{{10}}} \right)^0}{\left( {\dfrac{9}{8}} \right)^8} = 1 - {\left( {\dfrac{9}{8}} \right)^8}$
Hence the value of required ${\text{x = 8}}$

Note –Whenever we face such a type of problem statement the key concept that we need to recall is the concept of probability distribution of random variables .This helps to solve such a type of question and it will get you on the right track to reach the answer.