Probability question: At least 2 of 4?...at least 3 of 4?

Question

I have 4 cups side by side marked A, B, C and D. Each cup holds 4 identical marbles... 1 red and 3 blue. If I randomly select 1 marble from each cup what is the probability of picking out:
(a) at least 2 red marbles?
(b) at least 3 red marbles?

I am able to determine the probability of 0 red marbles (0.3164) , at least 1 red marble (0.6836) and 4 red marbles (0.0039) but have trouble in determining (a) and (b) listed above.

As a starting point I initially thought the answer for (a) lay with the probability of a red marble for Cup A (0.25) * at least 1 red marble for Cups B,C and D (0.5781)...which gives 0.1445. It seems overly simplistic to multiply this figure x 4 (for each cup) to give a total probability of 0.578 for at least 2 red marbles. This answer seems much too high.... given that at the probability of at least 1 red marble is 0.6836.

All help appreciated, especially the workings of how the answers to (a) and (b) are derived.

Answer 1

(a) You can think of the marble-picking as a sequence of n=4 Bernoulli trials. Success is choosing red, probability of success is p=0.25. The probability of k successes in n trials is

C(n,k)p^k (1-p)^(n-k)

where C(n,k) = n!/[k!(n-k)!] is the binomial coefficient. Use this with k=2, then 3, then 4. Add these probabilities together.

(b) In the process of doing (a), you'll get the answer to (b), just exclude the probability of exactly 2.

Answer 2

Since they are independent events, you multiply the probabilities.

Probability of red = 0.25
Probability of blue = 0.75

At least two red marbles = 2 red and 2 blue

= 0.25 x 0.25 x 0.75 x 0.75

At least three red marbles = 3 red and 1 blue

= 0.25 x 0.25 x 0.25 x 0.75

Hope that helps!