Question

a bag contains 26 marbles total, with 7 red marbles, 10 white marbles, and 9 blue marbles.
you draw 3 marbles out at random, without replacement. what is the probability that all 3 are red?
p(all 3 are red) =
what is the probability that 2 of the marbles are red, and 1 are not?
p(2 are red, 1 are not) =
what is the probability that none of the 3 marbles are red?
p(none are red) =
express your answers as fractions or a decimals accurate to at least 4 decimal places.

Explanation:

Step1: Calculate total ways to draw 3 marbles

Total marbles = 26. Number of ways to choose 3: $\binom{26}{3} = \frac{26!}{3!(26-3)!} = 2600$

Step2: Compute P(all 3 red)

Red marbles =7. Ways to choose 3 red: $\binom{7}{3}=35$. Probability: $\frac{35}{2600} \approx 0.0135$

Step3: Compute P(2 red,1 not red)

Non-red marbles=26-7=19. Ways: $\binom{7}{2} \times \binom{19}{1}=21×19=399$. Probability: $\frac{399}{2600}≈0.1535$

Step4: Compute P(none red)

Ways to choose 3 non-red: $\binom{19}{3}=969$. Probability: $\frac{969}{2600}≈0.3727$

Answer:

P(all 3 are red) ≈ 0.0135
P(2 are red, 1 are not) ≈ 0.1535
P(none are red) ≈ 0.3727