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Complex Probability Situations

Introduction

In real-world probability questions, we often deal with multiple stages, combined events, and interdependent outcomes - for example, drawing multiple balls with or without replacement, or combining outcomes from two or more experiments.

Understanding how to calculate probabilities in such mixed or multiple event scenarios is essential, as these problems combine conditional probability, independent/dependent events, and combinations of outcomes.

Pattern: Complex Probability Situations

Pattern

The key idea is to multiply probabilities for sequential events, adjusting for changing conditions (like without replacement) and summing probabilities when multiple pathways can lead to the same outcome.

Step-by-Step Example

Question

A box contains 3 red and 2 blue balls. Two balls are drawn one after another without replacement. What is the probability that both balls are red?

Solution

  1. Step 1: Find probability of first red ball

    There are 3 red out of 5 total balls, so P(R₁) = 3/5.

  2. Step 2: Find probability of second red ball

    After drawing one red, 2 red balls remain out of 4 total, so P(R₂ | R₁) = 2/4 = 1/2.

  3. Step 3: Multiply probabilities for both events

    Since draws are sequential and dependent: P(R₁ ∩ R₂) = (3/5) × (1/2) = 3/10.

  4. Final Answer:

    3/10.

  5. Quick Check:

    Favorable outcomes = 3C2 = 3; total outcomes = 5C2 = 10 → 3/10 ✅

Quick Variations

1. Drawing balls with replacement (independent events).

2. Drawing from multiple boxes or containers.

3. Problems involving at least one event happening.

4. Multi-stage experiments combining dice, cards, or urns.

Trick to Always Use

  • Step 1: Identify if the events are dependent or independent.
  • Step 2: Multiply sequential probabilities for “AND” cases.
  • Step 3: Add probabilities for “OR” cases or multiple pathways.
  • Step 4: Use combinations when choosing multiple items at once.

Summary

Summary

  • Complex probability situations involve multiple, often dependent, events.
  • For “AND” events → multiply probabilities.
  • For “OR” events → add probabilities (if mutually exclusive).
  • Always consider whether the problem involves replacement or no replacement.
  • Use combinations when dealing with multiple selections at once.

Practice

(1/5)
1. A box contains 4 red and 3 blue balls. Two balls are drawn one after another without replacement. What is the probability that both are red?
easy
A. 2/7
B. 3/10
C. 2/5
D. 4/7

Solution

  1. Step 1: Probability of first red

    P(R₁) = 4/7.

  2. Step 2: Probability of second red given first red

    P(R₂ | R₁) = 3/6 = 1/2.

  3. Step 3: Multiply sequential probabilities

    P(both red) = (4/7) × (1/2) = 2/7.

  4. Final Answer:

    2/7 → Option A.

  5. Quick Check:

    Favourable combinations = C(4,2)=6, total = C(7,2)=21 → 6/21 = 2/7 ✅
Hint: For 'both same color' without replacement, multiply sequential probabilities or use combinations.
Common Mistakes: Using replacement probabilities like (4/7)×(4/7).
2. A bag has 5 white and 4 black balls. Two balls are drawn together. Find the probability that both are black.
easy
A. 2/9
B. 1/6
C. 1/5
D. 2/7

Solution

  1. Step 1: Total ways to choose 2 balls

    Total = C(9,2) = 36.

  2. Step 2: Favourable ways (both black)

    Favourable = C(4,2) = 6.

  3. Step 3: Probability

    P = 6 / 36 = 1/6.

  4. Final Answer:

    1/6 → Option B.

  5. Quick Check:

    6 favourable out of 36 total → 1/6 ✅
Hint: When order doesn't matter, use combinations: favourable ÷ total.
Common Mistakes: Multiplying sequential fractions without accounting for unordered draw.
3. Two fair dice are thrown together. What is the probability that the sum is 7 or 11?
easy
A. 1/4
B. 2/11
C. 2/9
D. 1/6

Solution

  1. Step 1: Total ordered outcomes

    6 × 6 = 36.

  2. Step 2: Count favourable outcomes

    Sum 7 → 6 outcomes; Sum 11 → 2 outcomes; Total favourable = 8.

  3. Step 3: Probability

    P = 8 / 36 = 2/9.

  4. Final Answer:

    2/9 → Option C.

  5. Quick Check:

    8 out of 36 simplifies to 2/9 ✅
Hint: Add counts for each target sum (mutually exclusive events), then divide by 36.
Common Mistakes: Counting unordered pairs or double-counting outcomes.
4. A box contains 3 red, 4 blue, and 5 green balls. Two balls are drawn one after another without replacement. What is the probability that one is red and the other is blue?
medium
A. 1/6
B. 3/11
C. 12/55
D. 2/11

Solution

  1. Step 1: Total balls and total pairs

    Total = 12 balls; total pairs = C(12,2) = 66.

  2. Step 2: Favourable pairs (1 red & 1 blue)

    Favourable = C(3,1) × C(4,1) = 3 × 4 = 12.

  3. Step 3: Probability

    P = 12 / 66 = 2/11.

  4. Final Answer:

    2/11 → Option D.

  5. Quick Check:

    12 favourable out of 66 total → 2/11 ✅
Hint: For two different colours, multiply choices for each and divide by total combinations.
Common Mistakes: Using sequential fractions incorrectly or forgetting combinations count.
5. A bag contains 6 red, 5 blue, and 4 green balls. Two balls are drawn at random. Find the probability that both balls are of the same colour.
medium
A. 31/105
B. 13/30
C. 19/45
D. 11/30

Solution

  1. Step 1: Total ways to choose 2

    Total = C(15,2) = 105.

  2. Step 2: Favourable ways (same colour)

    Red: C(6,2)=15; Blue: C(5,2)=10; Green: C(4,2)=6. Sum = 15 + 10 + 6 = 31.

  3. Step 3: Probability

    P = 31 / 105 (cannot simplify) = 31/105.

  4. Final Answer:

    31/105 → Option A.

  5. Quick Check:

    31 favourable out of 105 total → 31/105 ✅
Hint: Add combinations for each colour and divide by total combinations.
Common Mistakes: Attempting to combine probabilities incorrectly instead of using combinations.

Mock Test

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