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Complex Probability Situations

Introduction

Real-world probability questions में अक्सर multiple stages, combined events और interdependent outcomes होते हैं - जैसे balls को replacement के साथ या बिना निकालना, या दो अलग-अलग experiments के outcomes को combine करना।

ऐसे mixed या multiple-event scenarios में probabilities निकालना ज़रूरी होता है, क्योंकि ये problems conditional probability, independent/dependent events और combinations of outcomes को एक साथ जोड़ती हैं।

Pattern: Complex Probability Situations

Pattern

मुख्य विचार यह है कि sequential events के लिए probabilities को multiply किया जाता है, और conditions बदलने पर (जैसे बिना replacement) probabilities को adjust किया जाता है। वहीं, जब एक outcome कई रास्तों से मिल सकता है, तो उन सभी probabilities को add किया जाता है।

Step-by-Step Example

Question

एक box में 3 red और 2 blue balls हैं। दो balls एक के बाद एक without replacement निकाली जाती हैं। Probability बताइए कि दोनों balls red हों।

Solution

  1. Step 1: पहली red ball की probability

    Total 5 balls में से 3 red हैं → P(R₁) = 3/5.

  2. Step 2: दूसरी red ball की probability

    एक red निकालने के बाद 2 red बचती हैं और total balls = 4 → P(R₂ | R₁) = 2/4 = 1/2.

  3. Step 3: दोनों events की joint probability

    Draws dependent हैं → 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. Replacement के साथ balls निकालना (independent events)।

2. Multiple boxes या containers से draw करना।

3. “At least one” वाले problems।

4. Dice, cards या urns वाले multi-stage experiments।

Trick to Always Use

  • Step 1: पहचानें कि events independent हैं या dependent।
  • Step 2: “AND” cases में sequential probabilities multiply करें।
  • Step 3: “OR” cases या multiple pathways में probabilities add करें।
  • Step 4: एक साथ multiple selections होने पर combinations का उपयोग करें।

Summary

Summary

  • Complex probability situations में कई events होते हैं, जो अक्सर dependent होते हैं।
  • “AND” events → probabilities multiply होती हैं।
  • “OR” events → probabilities add होती हैं (mutually exclusive होने पर)।
  • हमेशा देखें कि problem replacement के साथ है या बिना।
  • Multiple selections वाले cases में combinations का उपयोग करें।

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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