in a city, probability that a family has a cat is 0.2. P(X) = 0.2 find probability that in 10 families at least 2 of them has a cat.
$P(Y \geq 2) = ? $
I've tried to define a bernoulli distribution with paramenter 0.2. and then generalize it with a binomial distributiom with parameters (10, 0.2). Then, I rewrote $P(Y \geq 2)$ as $1 - P(Y \leq 2)$.
Now, $P(Y \leq 2)$ is equal to two binomial distributions solved for $k = 1$, and $k = 2$.
(for this question it's not important to solve this numerically, but only the logic behind a similar probability problem), I don't think it makes sense, because binomial distribution is discrete not continuous, and therefore I don't think $P(Y \geq 2)$ can be solved this way because in a discrete case distribution function (i.e $F_y()$) is not defined.
another way I was thinking of is to approximate the binomial distribution with a normal distribution using De Moivre-Laplace theorem. Of course, in order to use this theorem, I need to increase the sample size, otherwise I'd get a bad approximation of the final result.
I'll answer your question. A little suggestion: try to write in Latex.
The sought random variable is a binomial one, with parameters $n=10$ and $p = 0.2 = 1/5$, i.e. $X \sim \text{Bin}(10,1/5)$.
Now,
$$ \text{Pr}\left[X \ge 2\right] = 1 - \text{Pr}\left[X < 2\right] = 1 - \text{Pr}\left[X = 0 \right] - \text{Pr}\left[X = 1 \right]. $$
In general,
$$ \text{Pr}\left[X = k \right] = {n\choose k}p^k(1-p)^{n - k}. $$
Thus,
$$ \text{Pr}\left[X = 0\right]= {10\choose 0}\left({\frac 1 5}\right)^0\left({\frac 4 5}\right)^{10} = \left({\frac 4 5}\right)^{10}, $$
and
$$ \text{Pr}\left[X = 1\right]= {10\choose 1}\left({\frac 1 5}\right)^1\left({\frac 4 5}\right)^{9} = 2\left({\frac 4 5}\right)^{9}. $$
By plugging these results into the first formula, we get
$$ \text{Pr}\left[X \ge 2\right] = 1 - \left({\frac 4 5}\right)^{10} -2\left({\frac 4 5}\right)^{9} = 1 - \frac{4^{10} + \frac{5}{2}4^{10}}{5^{10}} = 1 - \frac{7}{2} \cdot \left(\frac{4}{5}\right)^{10} \approx 0.62 $$