Question
A bag contains 5 red and 7 blue balls. Two balls are drawn one after another without replacement. Find: (a) probability that both are red, (b) probability that the second is red given the first was red, (c) probability that exactly one is red. (CBSE pattern question — appears almost every year.)
Solution — Step by Step
Total balls = 12. Red balls = 5.
After drawing one red ball, 4 reds and 7 blues remain (11 total).
This is the conditional probability — the sample space has shrunk because we know the first was red.
Both red:
Exactly one red = (red then blue) + (blue then red):
(a) , (b) , (c) .
Why This Works
Conditional probability shrinks the sample space. Once you know an event has occurred, you only consider outcomes consistent with that event.
The multiplication rule chains conditional probabilities into joint probabilities. This is the bedrock of every “without replacement” problem.
The “exactly one” calculation requires summing over both orderings, since the question doesn’t specify which draw was red.
Whenever a problem says “without replacement”, expect conditional probability. Whenever it says “with replacement”, events are independent and you can multiply directly.
Alternative Method
Combinatorial approach for “both red”: . Same answer, but this method ignores the order — fine for “both” or “neither” but not for ordered conditional probabilities.
Common Mistake
For “exactly one red”, students often forget to add both orderings. If the draws are sequential and labelled (1st, 2nd), there are two ways to get exactly one red: RB or BR. Both have probability , summing to .