Question
A bag contains 5 red and 7 white balls. Two balls are drawn at random without replacement. Find the probability that (a) both are red, (b) one is red and one is white. A second student claims the answer to (b) is . What’s wrong with that?
Solution — Step by Step
Number of favourable: .
Choose 1 red from 5 and 1 white from 7: ways.
The expression equals — exactly half of the correct answer. That’s because the student computed the probability of “first red, then white” only. They forgot to add the case “first white, then red,” which has the same probability.
, .
Why This Works
When order doesn’t matter and we draw without replacement, two clean approaches exist: combinations (count favourable / total) or sequential probability (multiply, then sum over orders). Both give the same answer. The trap is in the second method — students forget that “one red and one white” includes both orderings.
The combinations route is safer for beginners because there’s no ordering to track — but it’s slower for problems with many cases.
Alternative Method
Sample space . Two reds: . Two whites: . One of each: . Same answer via complement.
Common Mistake
The student’s mistake — multiplying conditional probabilities without summing over orderings — is the single most common error in CBSE Class 12 probability. Whenever order doesn’t matter in the question but you used a sequential approach, double-check whether multiple orderings exist. For “one of each,” they always do.