Conditional Marble Draw with Three Colors
A bag contains $r$ red, $y$ yellow, and $b$ blue marbles, with $n = r + y + b$ total. You draw 2 marbles without replacement.
Given that at least one of the two drawn marbles is blue, what is the probability that neither marble is red?
Hints
- Frame this as $P(B \mid A)$ where $A$ is "at least one blue" and $B$ is "no red." Use the complement to count draws with at least one blue.
- The numerator counts draws from only $\{$yellow, blue$\}$ that include at least one blue: $\binom{y+b}{2} - \binom{y}{2}$.
- Expand the binomial coefficients and simplify. Check your formula with a small example like $r = y = b = 10$.
Worked Solution
How to Think About It: This is a conditional probability exercise -- you condition on seeing at least one blue, and want the chance that both marbles come from the non-red colors (blue and yellow). The key move is to set up the fraction correctly: numerator is the count of 2-marble draws that have at least one blue AND no red, denominator is all 2-marble draws with at least one blue. Both are easiest to compute via complements.
Quick Estimate: Suppose equal counts: $r = y = b = 10$, so $n = 30$. Total draws: $\binom{30}{2} = 435$. Draws with at least one blue: $435 - \binom{20}{2} = 435 - 190 = 245$. Draws with at least one blue and no red: draws from $\{y, b\}$ with at least one blue = $\binom{20}{2} - \binom{10}{2} = 190 - 45 = 145$. Probability: $145/245 = 29/49 \approx 0.592$.
Approach: Use the complement method for "at least one blue" events in both numerator and denominator.
Formal Solution:
Let $A$ = "at least one blue" and $B$ = "no red marble drawn." We want $P(B \mid A)$.
$$P(B \mid A) = \frac{P(A \cap B)}{P(A)}$$
Denominator -- $P(A)$: Probability of at least one blue.
$$P(A) = 1 - \frac{\binom{n-b}{2}}{\binom{n}{2}} = \frac{\binom{n}{2} - \binom{r+y}{2}}{\binom{n}{2}}$$
Numerator -- $P(A \cap B)$: Both marbles are from $\{y, b\}$ (no red) AND at least one is blue.
$$P(A \cap B) = \frac{\binom{y+b}{2} - \binom{y}{2}}{\binom{n}{2}}$$
The subtraction removes the case where both are yellow (non-red but also non-blue).
So:
$$P(B \mid A) = \frac{\binom{y+b}{2} - \binom{y}{2}}{\binom{n}{2} - \binom{r+y}{2}}$$
Expanding the binomial coefficients:
$$= \frac{(y+b)(y+b-1) - y(y-1)}{n(n-1) - (r+y)(r+y-1)}$$
$$= \frac{2by + b(b-1)}{2bn - b(b-1) - 2b \cdot 0}$$
Simplifying the numerator: $(y+b)(y+b-1) - y(y-1) = b^2 + 2by - b = b(b + 2y - 1)$.
Simplifying the denominator: $n(n-1) - (n-b)(n-b-1) = 2b(n-b) - b + b = b(2n - 2b - 1 + 1) = b(2n - b - 1)$... let's be more careful:
$n(n-1) - (r+y)(r+y-1) = n^2 - n - (n-b)^2 + (n-b) = 2nb - b^2 - b = b(2n - b - 1)$.
So:
$$P(B \mid A) = \frac{b(b + 2y - 1)}{b(2n - b - 1)} = \frac{b + 2y - 1}{2n - b - 1}$$
Verification with equal counts ($r = y = b = 10$, $n = 30$): $(10 + 20 - 1)/(60 - 10 - 1) = 29/49 \approx 0.592$. This matches the quick estimate.
Answer: $P(\text{no red} \mid \text{at least one blue}) = \dfrac{b + 2y - 1}{2n - b - 1}$, where $n = r + y + b$. The answer depends on the composition of the bag; with equal counts ($r = y = b$), it equals $(3b - 1)/(5b - 1)$.
Intuition
This is a classic conditional probability problem that tests whether you can carefully set up a ratio of counting arguments. The subtlety is that "one of them is blue" means at least one blue, not exactly one -- and the complement method (total minus the unwanted) is cleaner than direct enumeration for both numerator and denominator.
A common mistake is to assume equal counts and give a single number. In an interview, the right move is to note that the answer depends on the bag composition, write the general formula, then plug in a specific example to sanity-check. This shows you understand the structure rather than just guessing. The problem also illustrates why you should always verify conditional probability setups with extreme cases: if $r = 0$ the answer should be 1 (no red marbles exist), and indeed $(b + 2y - 1)/(2(y+b) - b - 1) = (b + 2y - 1)/(2y + b - 1) = 1$. Good.