What is the probability if I draw $1$ of $100$ different balls that this $1$ is one of $3$ predefined ones?

Question

What is the probability if I draw $1$ of $100$ different balls that this drawn ball is one of $3$ predefined ones?

Is it $3/100$, right?

And what is the probability of obtaining at least one of the $3$ predefined ones when drawing $2$ balls out of $100$ without replacement?

Is it $2 \cdot 3/100$, right?

Answer 1

1. Yes. Three choices out of 100 to draw one of the three predefined ones.

2. No. ${3 \choose 2}$ ways to select the predefined ones. ${100\choose 2}$ ways to select any two balls.

$$\frac{3}{\frac{100\times 99}{2}}=\frac{6}{9900}$$

ways to draw two balls of which turn out to be two of the three predefined ones. This is about $\frac 2{3300}=\frac 1{1650}$

---

If at least one of the two balls drawn is one of the predefined ones, it's

$$1-\frac{{97\choose 2}}{{100\choose 2}}=1-\frac{\frac{97\times96}{2}}{\frac{100\times99}{2}}=1-\frac{97\times48}{50\times99}=1-\frac{97\times8}{25\times33}$$

Answer 2

1. Yes. You have $3/100$ chance of selecting one of the predefined balls. Because choosing one of $m$ balls from a population of $n$ balls is $m/n$.
2. No. You have $3/100$ chance of selecting one of the predefined balls in your first draw (the same as before). If you don't get one in the first draw (with probability $97/100$), then you have $3/99$ (choosing one of $3$ balls from the remaining $99$) chance of selecting one in your second draw. So total probability of getting at least one of the predefined balls in two draws is $3/100 + (97/100)*(3/99) = 3/100*(1+97/99)$ which is almost $6/100$.