Question
A box contains red balls and blue balls. Two balls are drawn without replacement. Given that the second ball is red, find the probability that the first ball was also red.
Solution — Step by Step
Let = “first ball is red,” = “second ball is red.” We need .
By Bayes’ theorem:
(5 red out of 8 total). (after drawing red first, 4 red out of 7 remain).
Final answer: .
Why This Works
Conditional probability flips the direction of inference. We know — the forward direction. To get — the backward inference — we use Bayes’ theorem.
A surprising result: . The probability of red on any single draw (without conditioning) is just the original proportion of red balls. This is the “exchangeability” of draws without replacement.
Alternative Method
Direct counting: total ordered pairs where second is red = (cases where first is also red) + (cases where first is blue). First red, second red: . First blue, second red: . Total: . Of these, the favourable ones (first red AND second red) = . So . Same answer, cleaner if you’re comfortable with counting.
Students often answer , treating . That assumes independence — but draws without replacement are NOT independent. The condition does change the answer, just to , not .
Common Mistake
A subtle trap: assuming . These are different in general (Bayes’ theorem connects them via prior probabilities). Only when does the equality hold — which happens here, but it’s coincidence, not a general rule.