Question
A bag contains 5 red and 3 white balls. Two balls are drawn one after the other without replacement. Find the probability that:
(a) Both are red. (b) The second ball is red. (c) The first ball was red, given that the second is red.
Solution — Step by Step
The probability of red on the first draw is . After one red is removed, 4 reds remain in 7 balls, so the conditional probability of red on the second draw is .
Condition on the first draw. The first ball is either red or white:
The second-draw probability equals the first-draw probability — a beautiful symmetry of without-replacement drawing.
We want :
Final answers: (a) , (b) , (c) .
Why This Works
The “second ball red” probability matches the “first ball red” probability because, when no information about the order is used, every ball is equally likely to be in any position. This exchangeability is a powerful shortcut for JEE Main MCQs.
Part (c) asks the reverse of the natural draw direction — given that the second is red, what is the chance the first was red? Bayes’ theorem reframes the question using the joint probability we already computed in (a).
Alternative Method
For part (b), use exchangeability directly: any ball drawn, in any position, is red with probability equal to (red count)/(total) = . No conditioning needed.
For part (c), use Bayes in unsimplified form:
Common Mistake
Students assume because the second draw “depends on” the first. But the unconditional probability is the same — only the conditional probabilities given a specific first outcome differ. Exchangeability is one of those topics that costs students 2 to 3 marks every paper.