Probability: Numerical Problems Set (9)

hard • 3 min read
Tags Probability

Question

A bag contains 44 red and 66 blue balls. Two balls are drawn one after another without replacement. Find (a) the probability both are red, (b) the probability the second is blue given the first is red, (c) the probability of one red and one blue (in any order). Then redo the entire problem with replacement and explain why the answers differ.

Solution — Step by Step

P(1st red)=410=25P(\text{1st red}) = \frac{4}{10} = \frac{2}{5}

After removing one red, 33 red remain in 99 balls:

P(2nd red∣1st red)=39=13P(\text{2nd red} \mid \text{1st red}) = \frac{3}{9} = \frac{1}{3}

By multiplication rule:

P(both red)=25×13=215P(\text{both red}) = \frac{2}{5} \times \frac{1}{3} = \frac{2}{15}

After removing one red, all 66 blue still in the bag of 99:

P(2nd blue∣1st red)=69=23P(\text{2nd blue} \mid \text{1st red}) = \frac{6}{9} = \frac{2}{3}

Two cases:

  • P(R then B)=410×69=2490=415P(\text{R then B}) = \tfrac{4}{10} \times \tfrac{6}{9} = \tfrac{24}{90} = \tfrac{4}{15}
  • P(B then R)=610×49=2490=415P(\text{B then R}) = \tfrac{6}{10} \times \tfrac{4}{9} = \tfrac{24}{90} = \tfrac{4}{15}

Total: 415+415=815\tfrac{4}{15} + \tfrac{4}{15} = \tfrac{8}{15}.

When the first ball is replaced, the bag returns to its original state.

  • P(both red)=410×410=425P(\text{both red}) = \tfrac{4}{10} \times \tfrac{4}{10} = \tfrac{4}{25}
  • P(2nd blue∣1st red)=610=35P(\text{2nd blue} \mid \text{1st red}) = \tfrac{6}{10} = \tfrac{3}{5} (independence)
  • P(one R one B)=2×410×610=48100=1225P(\text{one R one B}) = 2 \times \tfrac{4}{10} \times \tfrac{6}{10} = \tfrac{48}{100} = \tfrac{12}{25}

Why This Works

The crucial distinction is independence vs dependence. With replacement, the two draws are independent: the second draw’s probabilities don’t depend on the first. Without replacement, drawing a red ball changes the composition for the second draw — events are dependent, and we need conditional probability.

The multiplication rule P(A∩B)=P(A)⋅P(B∣A)P(A \cap B) = P(A) \cdot P(B \mid A) handles both cases. For independent events, P(B∣A)=P(B)P(B \mid A) = P(B), and the rule reduces to P(A)⋅P(B)P(A) \cdot P(B).

Alternative Method

For “without replacement” problems, you can sometimes count ordered pairs directly using combinations. Total ways to pick 2 balls from 10: (102)=45\binom{10}{2} = 45. Both red: (42)=6\binom{4}{2} = 6. So P(both red)=6/45=2/15P(\text{both red}) = 6/45 = 2/15. ✓

For one of each: (41)(61)=24\binom{4}{1}\binom{6}{1} = 24. P=24/45=8/15P = 24/45 = 8/15. ✓

This combinatorial approach is faster when the order doesn’t matter for the question.

Always identify replacement vs no-replacement before writing any equation. The single most-asked variation in JEE/CBSE probability is to flip this assumption mid-problem to test whether you noticed.

Common Mistake

Two standard errors:

  1. Forgetting to count both orders for “one of each”. The event “one red, one blue” includes both R-then-B and B-then-R unless the problem specifies an order. Doubling is needed in the without-replacement case (each order has the same probability here, but they’re different events).

  2. Using replacement probabilities in a no-replacement problem. If you compute P(both red)P(\text{both red}) as 4/10×4/104/10 \times 4/10 when it should be 4/10×3/94/10 \times 3/9, you’ll be off by a factor of ∼1.5\sim 1.5. Always check whether the bag is restored.

Final answer: Without replacement: P(both red)=2/15P(\text{both red}) = 2/15, P(2nd blue∣1st red)=2/3P(\text{2nd blue} \mid \text{1st red}) = 2/3, P(one each)=8/15P(\text{one each}) = 8/15. With replacement: 4/254/25, 3/53/5, 12/2512/25.

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