Question
A bag contains 4 red and 6 black balls. Two balls are drawn one after the other without replacement. Find the probability that (a) both are red, (b) the first is red and the second is black, (c) at least one is red.
Solution — Step by Step
“Without replacement” means the second draw depends on the first. We will use the multiplication rule: P(A∩B)=P(A)⋅P(B∣A).
P(first red)=4/10=2/5. After one red is removed, 3 red and 6 black remain. P(second red | first red)=3/9=1/3.
P(both red)=52⋅31=152
P(first red)=4/10. Then 3 red and 6 black remain. P(second black)=6/9=2/3.
P(R then B)=104⋅96=9024=154
Use the complement: P(at least one red)=1−P(both black).
P(both black)=106⋅95=9030=31.
P(at least one red)=1−31=32
Final Answer: (a) 2/15, (b) 4/15, (c) 2/3.
Why This Works
Without replacement, the sample space shrinks after each draw, which is why P(second red | first red)=P(first red). The multiplication rule formalizes this dependence.
For “at least one” questions, the complement trick saves time. Counting “both black” is one calculation; counting “exactly one red” plus “exactly two reds” is two calculations. Always check if the complement is easier.
Alternative Method
Use combinations: total ways to pick 2 from 10 is (210)=45. Both red: (24)=6 ways. So P=6/45=2/15. Quicker for unordered draws, but doesn’t directly handle ordered pairs like part (b).
Using P(second red)=4/10 in part (a) treats the draws as independent. They’re not — once a red is drawn, only 3 reds remain in 9 balls. Always check whether the question says “with” or “without” replacement.
The complement rule P(at least one)=1−P(none) is your fastest path for “at least” questions. Train this reflex — JEE Main asks at least one such question every year.