A box contains 7 blue pens and 5 red pens. Four pens are randomly selected from the box. We want to calculate the probability of two events: A: All 4 pens are the same color. B: At most 2 red pens are selected.

1. Problem Description

A box contains 7 blue pens and 5 red pens. Four pens are randomly selected from the box. We want to calculate the probability of two events:

A: All 4 pens are the same color.

B: At most 2 red pens are selected.

2. Solution Steps

A: All 4 pens are the same color.

This means either all 4 are blue or all 4 are red.

The total number of ways to choose 4 pens from 12 is $\binom{12}{4} = \frac{12!}{4!8!} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495$.

The number of ways to choose 4 blue pens from 7 is $\binom{7}{4} = \frac{7!}{4!3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$.

The number of ways to choose 4 red pens from 5 is $\binom{5}{4} = \frac{5!}{4!1!} = 5$.

So, the number of ways to choose 4 pens of the same color is $35 + 5 = 40$.

The probability of event A is $P(A) = \frac{40}{495} = \frac{8}{99}$.

B: At most 2 red pens are selected.

This means we can have 0, 1, or 2 red pens. It also means 4, 3, or 2 blue pens are selected respectively.

Number of ways to choose 0 red pens and 4 blue pens: $\binom{5}{0} \binom{7}{4} = 1 \times 35 = 35$.

Number of ways to choose 1 red pen and 3 blue pens: $\binom{5}{1} \binom{7}{3} = 5 \times \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 5 \times 35 = 175$.

Number of ways to choose 2 red pens and 2 blue pens: $\binom{5}{2} \binom{7}{2} = \frac{5 \times 4}{2 \times 1} \times \frac{7 \times 6}{2 \times 1} = 10 \times 21 = 210$.

So, the number of ways to choose at most 2 red pens is $35 + 175 + 210 = 420$.

The probability of event B is $P(B) = \frac{420}{495} = \frac{28}{33}$.

3. Final Answer

A: The probability that all 4 pens are the same color is $\frac{8}{99}$.

B: The probability that at most 2 red pens are selected is $\frac{28}{33}$.