Same-Color Pair After Drawing
You have an urn containing 7 red balls and 8 blue balls (15 total). You draw balls one at a time without replacement until only two balls remain in the urn.
What is the probability that the two remaining balls are the same color?
Hints
- Think about what the "last two balls" really means -- does the order of drawing matter?
- Drawing without replacement is exchangeable. The last two balls form a uniformly random pair from the original 15.
- Count same-color pairs: $\binom{7}{2} + \binom{8}{2}$, and divide by the total number of pairs $\binom{15}{2}$.
Worked Solution
How to Think About It: The key insight here is that drawing without replacement is exchangeable -- the order in which you draw doesn't matter. The two balls left at the end are just a uniformly random pair from the original 15. So you don't need to think about the sequential drawing process at all. You're simply asking: if I pick 2 balls at random from the urn, what's the chance they match in color? This reframing makes the problem trivial.
Quick Estimate: There are 7 red and 8 blue. Two reds is roughly $7/15 \times 6/14 \approx 0.47 \times 0.43 \approx 0.20$. Two blues is roughly $8/15 \times 7/14 \approx 0.53 \times 0.50 \approx 0.27$. Sum is about $0.47$, so the answer should be just under
Approach: Use combinations to count the favorable pairs over total pairs.
Formal Solution:
Total ways to choose the remaining 2 balls from 15:
$\binom{15}{2} = 105$
Same-color pairs:
$\binom{7}{2} + \binom{8}{2} = 21 + 28 = 49$
So the probability is:
$P(\text{same color}) = \frac{49}{105} = \frac{7}{15}$
Answer: The probability that the two remaining balls are the same color is $\dfrac{7}{15} \approx 0.467$.
Intuition
The deep lesson here is exchangeability: in a random permutation, any subset of positions is equally likely to hold any subset of items. When you draw without replacement, the last two balls are distributed identically to the first two balls you would draw. This means you can ignore the entire sequential process and just compute a simple combinatorial ratio. This principle shows up constantly in quantitative interviews -- problems that look like they require tracking a complicated sequential process often collapse to a simple counting argument once you recognize exchangeability. The common mistake is trying to condition on each draw step by step, which gives the same answer but with far more work.