How Many Green Gloves in the Box?
A box contains 3 blue gloves and $n$ green gloves. You draw two gloves from the box without replacement. If the probability of drawing 2 blue gloves is $\frac{1}{2}$, what is the value of $n$?
Hints
- Write the probability of drawing 2 blue gloves from $3 + n$ total gloves as a product of sequential probabilities.
- Set $\frac{3}{3+n} \cdot \frac{2}{2+n} = \frac{1}{2}$ and cross-multiply to get a quadratic in $n$.
- The quadratic $n^2 + 5n - 6 = 0$ factors as $(n+6)(n-1) = 0$. Only the positive root is valid.
Worked Solution
How to Think About It: This is a hypergeometric probability setup. Two gloves drawn without replacement from a box with 3 blue and $n$ green. We want $P(\text{both blue}) = 1/2$. That probability is $\frac{\binom{3}{2}}{\binom{3+n}{2}}$. Set it equal to $1/2$ and solve for $n$. Before computing, note that $P = 1/2$ is very high for drawing two blue gloves, so $n$ must be very small -- probably 0 or 1.
Quick Estimate: With $n = 1$ (4 gloves total): $P = \frac{3}{4} \cdot \frac{2}{3} = \frac{6}{12} = \frac{1}{2}$. That works. With $n = 0$ (3 gloves): $P = 1$. With $n = 2$ (5 gloves): $P = \frac{3}{5} \cdot \frac{2}{4} = \frac{6}{20} = 0.3$. So $n = 1$ is the unique answer.
Approach: Set up the probability equation and solve.
Formal Solution:
The probability of drawing 2 blue gloves without replacement from a box of $3 + n$ gloves (3 blue, $n$ green) is:
$$P(\text{both blue}) = \frac{3}{3+n} \cdot \frac{2}{2+n}$$
Setting this equal to $1/2$:
$$\frac{3}{3+n} \cdot \frac{2}{2+n} = \frac{1}{2}$$
$$\frac{6}{(3+n)(2+n)} = \frac{1}{2}$$
$$(3+n)(2+n) = 12$$
$$n^2 + 5n + 6 = 12$$
$$n^2 + 5n - 6 = 0$$
$$(n+6)(n-1) = 0$$
So $n = 1$ or $n = -6$. Since $n$ must be a non-negative integer, $n = 1$.
Verification: With $n = 1$, the box has 4 gloves (3 blue, 1 green). $P(\text{both blue}) = \frac{3}{4} \cdot \frac{2}{3} = \frac{1}{2}$. Correct.
Answer: $n = 1$.
Intuition
This is a classic "work backwards from the probability" problem. You are given the answer ($P = 1/2$) and asked to find the parameter ($n$). The hypergeometric setup (sampling without replacement) gives a rational function of $n$, which reduces to a quadratic. The key insight is that $P = 1/2$ is surprisingly high, which immediately tells you $n$ must be very small -- there can be at most 1 green glove for the blue-pair probability to be that large.
In practice, this type of reasoning shows up when calibrating models. If you observe a certain empirical frequency and want to infer the population parameter, you are doing exactly this -- inverting a probability formula. The quadratic structure here is simple, but the same logic extends to maximum likelihood estimation in more complex settings.