Probability: Step-by-Step Worked Examples (2)

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Question

A bag contains 4 red, 5 blue, and 6 green balls. Three balls are drawn at random without replacement. What is the probability that exactly two are blue?

Solution — Step by Step

Total balls = 15. Total ways:

(153)=15!3!12!=455\binom{15}{3} = \frac{15!}{3! \cdot 12!} = 455

Exactly 2 blue means: choose 2 blue from 5, and 1 non-blue from the remaining 10 (4 red + 6 green).

(52)×(101)=10×10=100\binom{5}{2} \times \binom{10}{1} = 10 \times 10 = 100

P=100455=2091P = \frac{100}{455} = \frac{20}{91}

Final answer: P=2091P = \dfrac{20}{91}

Why This Works

When draws are without replacement and order doesn’t matter, combinations (not permutations) count outcomes correctly. The formula P=favourable combinationstotal combinationsP = \dfrac{\text{favourable combinations}}{\text{total combinations}} relies on every combination being equally likely.

The “exactly 2 blue” requirement means we deliberately avoid the third blue — that’s why we draw the third from the non-blue pool.

Alternative Method

Use the multiplication rule on ordered draws and then divide by the number of orderings. P(BBN) for one specific order = (5/15)(4/14)(10/13)(5/15)(4/14)(10/13). There are (32)=3\binom{3}{2} = 3 orderings of where the non-blue lands. Multiply: 3×(5/15)(4/14)(10/13)=600/2730=20/913 \times (5/15)(4/14)(10/13) = 600/2730 = 20/91. Same answer.

Common Mistake

Students compute “at least 2 blue” instead of “exactly 2 blue”. “At least 2” includes the case of all 3 blue, which adds (53)=10\binom{5}{3} = 10 favourable outcomes. The two answers differ — read the question carefully.

For “without replacement, count by combinations” is the default approach in CBSE. For “with replacement” or “ordered”, use multiplication of independent probabilities.

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