抽取3个球的颜色组合概率计算及结果验证咨询

Let's work through your two probability questions and fix the mistake in your reasoning for the second one.

Question 1: Probability of drawing 2 blue balls and 1 red ball in 3 draws #

To calculate this accurately, we need a bit more context:

• Are the draws with replacement (you put the ball back after each draw) or without replacement?
• What's the total number of balls in the urn, or the fixed probability of drawing a blue/red ball on a single trial?

If we assume draws are with replacement, and let ( p_b ) = probability of drawing a blue ball, ( p_r ) = probability of drawing a red ball, the formula is:
$$
\binom{3}{2} \times p_b^2 \times p_r = 3p_b^2p_r
$$
For without replacement, if there are ( B ) blue balls, ( R ) red balls, and ( N ) total balls in the urn, the formula becomes:
$$
\frac{\binom{B}{2} \times \binom{R}{1}}{\binom{N}{3}}
$$

Question 2: Probability of getting at least 1 blue and 1 red in 3 draws (with ( p_b=0.4 ), ( p_r=0.25 )) #

Your initial calculation has two critical errors:

1. You only accounted for one valid scenario: "At least 1 blue and 1 red" covers three distinct outcomes, not just 1 blue, 1 red, and 1 non-blue/non-red ball. The full set of valid outcomes is:
   • 1 blue, 1 red, 1 non-blue/non-red (let's call this ( p_o = 1 - 0.4 - 0.25 = 0.35 ))
   • 2 blue, 1 red
   • 1 blue, 2 red
2. Your formula omitted the non-blue/non-red probability: For the 1B1R1O case, you need to multiply by ( p_o ) — your initial calculation only used ( 0.4 \times 0.25 ), which misses this key term.

Let's calculate each valid scenario step by step:

• 1B1R1O: The number of distinct permutations is ( \frac{3!}{1!1!1!} = 6 ). Probability: ( 6 \times 0.4 \times 0.25 \times 0.35 = 0.21 )
• 2B1R: The number of ways to choose which 2 draws are blue is ( \binom{3}{2} = 3 ). Probability: ( 3 \times (0.4)^2 \times 0.25 = 0.12 )
• 1B2R: The number of ways to choose which 1 draw is blue is ( \binom{3}{1} = 3 ). Probability: ( 3 \times 0.4 \times (0.25)^2 = 0.075 )

Adding these probabilities together gives the total valid probability:
$$
0.21 + 0.12 + 0.075 = 0.405
$$
This aligns almost perfectly with your Excel simulation result (~0.4) — your initial calculation was too high because it missed most of the valid outcomes and ignored the non-blue/non-red ball probability.

内容的提问来源于stack exchange，提问作者monsterhaij