Question
A bag contains 5 red and 3 blue balls. Two balls are drawn without replacement. Find the probability that both are red.
Solution — Step by Step
Without replacement → conditional probability or counting. Both work; counting is faster here.
Total ways to choose 2 balls from 8: .
Favourable ways (both red, choose 2 from 5): .
. After drawing one red, .
. Same answer.
Probability .
Why This Works
The combination method counts unordered pairs — we don’t care about which ball came first. The sequential method tracks order but multiplies probabilities — both approaches lead to the same answer because the unordering and the conditioning balance out.
For board exams, the sequential method is often easier to write up. For JEE Main MCQs, the combination method is faster.
Speed shortcut: For “without replacement” problems, where is favourable count, is total, is draws. One formula for any number of balls.
Alternative Method — Conditional Probability Tree
Draw the tree: first draw red () or blue (). After red, branch into red () or blue (). After blue, into red () or blue ().
The “both red” branch: .
This method scales easily to “at least one red” or “exactly one red” follow-ups.
Common Mistake
Students often compute , treating draws as independent. That formula applies to WITH REPLACEMENT, not without.
Another classic: using — mismatching the two binomials. Always use the same in both numerator and denominator.
JEE Main 2023 (Shift 2, April 8) had a near-identical structure with 6 white and 4 black, drawing 3 without replacement. The pattern is so common that recognising “without replacement” should immediately trigger the formula. CBSE Class 12 boards ask this every alternate year as a 4-mark question.
For NEET, basic probability isn’t tested — but the underlying logic shows up in genetics problems (Punnett squares), so the foundation is still useful.