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  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?

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No, though you were close for the second one. Please try again.

Hint 1: what's the probability that the first ball is red? If given that, what's the probability that the second ball is blue? Put it together.

1. : 3 of the 10 balls we could draw first are red, and after drawing once, 2 of the remaining 8 balls we can draw second are blue.

Hint 2: You were fairly close. Think about what you are selecting and from where.

2. : Count the ways to pick 2 of the 5 red balls and 2 of the 7 blue balls; out of all the ways to pick 4 of the 12 balls.

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