# Questions that make use of the naive definition of probability

1. We have 5 balls, 3 red and 2 blue. If we where to arrange these balls in a row randomly what is the naive probability that the first ball is red and second ball is blue?

The way I approached this question is by using the naive definition of probability which says that,

$$P(A) = \frac{\# \text{ of favourable outcomes}}{\#\text{ of possible outcomes}}$$

Let A be the event that the first ball is red and the second ball is blue.

number of possible outcomes = $\frac{5!}{3!2!}$

number of favourable outcomes = 3 choices(from the 3 reds) for the first ball, 2 choices(from the 2 blues) for the second ball, then the remaining balls can be in any order so 3!. Thus, the number of favourable outcomes is $3 \cdot 2 \cdot 3 \cdot 2 \cdot 1$

therefore, $$P(A) = \frac{3 \cdot 2 \cdot 3 \cdot 2 \cdot 1}{\frac{5!}{3!2!}}$$

1. Suppose we have a pot with 5 red balls and 7 blue balls inside. If we were to pick 4 balls out of the pot without replacement what is the naive probability that 2 of the balls we picked are red?

Using the same idea as before,

Let A be the event that two of the balls we picked are red.

number of possible outcomes = ${12 \choose 4}$ because we are picking 4 out of the 12 total balls.

number of favourable outcomes = we choose 2 reds from the 5 total reds(${5 \choose 2}$) and we pick 2 from the remaining(${10 \choose 2}$). So, ${5 \choose 2} \cdot {10 \choose 2}$

therefore,

$$P(A) = \frac{{5 \choose 2} \cdot {10 \choose 2}}{{12 \choose 2}}$$

Is this the correct way to approach these questions? and are my solutions to these problem correct?