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Conditional Probability

Introduction

Conditional Probability किसी event के होने की probability को measure करता है जब यह पता हो कि कोई दूसरा event पहले ही हो चुका है। यह probability का एक बहुत महत्वपूर्ण concept है और real-world contexts जैसे weather forecasting, quality control और risk analysis में उपयोग होता है।

मुख्य विचार यह है कि sample space को restrict किया जाता है - केवल वही outcomes consider किए जाते हैं जिनमें पहला event हो चुका हो, और उसी narrowed-down space में दूसरा event evaluate किया जाता है।

Pattern: Conditional Probability

Pattern

Event B की probability, यह जानते हुए कि event A हो चुका है, इस प्रकार दी जाती है:

P(B | A) = P(A ∩ B) / P(A)
यह formula किसी ऐसे event की probability निकालने में मदद करता है जिसकी calculation दूसरे related event के already हो जाने पर dependent होती है।

Step-by-Step Example

Question

52 cards की standard deck से एक card निकाला जाता है। Probability बताएं कि वह King हो, यह जानते हुए कि वह एक face card है।

Solution

  1. Step 1: Known information पहचानें

    Total cards = 52. Face cards = {J, Q, K} चारों suits से → कुल 12 face cards।
  2. Step 2: Favourable cases पहचानें

    Face cards में Kings = 4 (प्रत्येक suit में एक King)।
  3. Step 3: Conditional probability formula apply करें

    P(King | Face card) = P(King ∩ Face card) ÷ P(Face card) = (4/52) ÷ (12/52) = 4/12 = 1/3.
  4. Final Answer:

    1/3.
  5. Quick Check:

    12 face cards में से 4 Kings हैं → 4/12 = 1/3 ✅

Quick Variations

1. Face card होने की condition पर red card की probability।

2. Student के Math pass करने की probability, यह जानते हुए कि उसने Science pass किया है।

3. किसी specific machine से चुने गए item के defective होने की probability।

Trick to Always Use

  • Step 1: Sample space को उस condition तक narrow करें जो “given” है।
  • Step 2: Formula P(B | A) = P(A ∩ B) / P(A) का उपयोग करें।
  • Step 3: “Given that” हमेशा नया denominator तय करता है - इस बात पर ध्यान दें।

Summary

Summary

Conditional Probability pattern में:

  • Formula: P(B | A) = P(A ∩ B) / P(A).
  • “Given” event (A) नया sample space बन जाता है।
  • यह तब उपयोग होता है जब events आपस में related हों; जरूरी नहीं कि वे independent हों।
  • हमेशा सुनिश्चित करें कि P(A) ≠ 0 हो (impossible event पर condition नहीं लगा सकते)।

Practice

(1/5)
1. A single card is drawn from a well-shuffled 52-card deck. Given that the card is a Heart, what is the probability that it is an Ace?
easy
A. 1/13
B. 1/4
C. 1/26
D. 1/52

Solution

  1. Step 1: Restrict the sample space

    If the card is a Heart, the sample space is the 13 Heart cards.
  2. Step 2: Count favourable outcomes

    There is exactly 1 Ace of Hearts among the 13 Hearts.
  3. Step 3: Apply conditional probability

    P(Ace | Heart) = 1 / 13 = 1/13.
  4. Final Answer:

    1/13 → Option A.
  5. Quick Check:

    Out of 13 Hearts only one is an Ace → 1/13 ✅
Hint: When given a suit, use only that suit's 13 cards as the denominator.
Common Mistakes: Using 52 as denominator instead of the conditional 13.
2. Two fair dice are rolled. Given that at least one die shows 4, what is the probability that the sum of the dice is 7?
easy
A. 2/11
B. 1/6
C. 1/5
D. 3/11

Solution

  1. Step 1: Build the conditional sample space

    Ordered outcomes with at least one 4: (4,1..6) and (1..6,4) minus (4,4) counted twice → 6 + 6 - 1 = 11 outcomes.
  2. Step 2: Find favourable outcomes

    Ordered pairs that sum to 7 and include a 4 are (3,4) and (4,3) → 2 favourable outcomes.
  3. Step 3: Compute conditional probability

    P(sum = 7 | at least one 4) = 2 / 11 = 2/11.
  4. Final Answer:

    2/11 → Option A.
  5. Quick Check:

    Total conditional outcomes = 11 and only two give sum 7 → 2/11 ✅
Hint: List ordered pairs for the condition first, then count favourable ones.
Common Mistakes: Treating unordered pairs or forgetting to subtract the double-counted (4,4).
3. In a class, 60% of students passed Mathematics, 50% passed Science, and 30% passed both subjects. Given that a student passed Mathematics, what is the probability that they also passed Science?
easy
A. 1/3
B. 1/2
C. 3/5
D. 2/3

Solution

  1. Step 1: Identify given probabilities

    P(Math) = 0.60, P(Math ∩ Science) = 0.30.
  2. Step 2: Use conditional formula

    P(Science | Math) = P(Math ∩ Science) / P(Math) = 0.30 / 0.60 = 0.5.
  3. Final Answer:

    1/2 → Option B.
  4. Quick Check:

    Half of the Math-passers also passed Science since 30% out of 60% = 1/2 ✅
Hint: Divide the intersection probability by the given event's probability.
Common Mistakes: Using P(Science) instead of P(Math ∩ Science) in the numerator.
4. Two cards are drawn successively from a well-shuffled 52-card deck without replacement. Given that the first card drawn is an Ace, what is the probability that the second card is also an Ace?
medium
A. 1/13
B. 1/17
C. 3/51
D. 1/52

Solution

  1. Step 1: Recognise the condition

    The first card is known to be an Ace, so one Ace has been removed from the deck.
  2. Step 2: Count remaining favourable and total cards

    Remaining Aces = 3, remaining cards = 51 → P = 3/51.
  3. Step 3: Simplify (optional)

    3/51 simplifies to 1/17, but the conditional probability directly is 3/51.
  4. Final Answer:

    3/51 → Option C.
  5. Quick Check:

    After one Ace is drawn there are 3 Aces left out of 51 cards → 3/51 ✅
Hint: After removing a known card, update counts for the second draw: favourable/remaining total.
Common Mistakes: Keeping denominator 52 (forgetting card was not replaced).
5. Box A contains 3 red and 2 blue balls. Box B contains 1 red and 4 blue balls. A box is chosen at random and one ball is drawn; it is red. What is the probability that the ball was drawn from Box A?
medium
A. 3/5
B. 1/3
C. 1/4
D. 3/4

Solution

  1. Step 1: Write known probabilities

    P(A) = 1/2, P(B) = 1/2. P(red | A) = 3/5. P(red | B) = 1/5.
  2. Step 2: Use Bayes' rule

    P(A | red) = P(red | A)P(A) / [P(red | A)P(A) + P(red | B)P(B)].
  3. Step 3: Plug in numbers

    = (3/5 × 1/2) / (3/5 × 1/2 + 1/5 × 1/2) = (3/10) / (3/10 + 1/10) = (3/10) / (4/10) = 3/4.
  4. Final Answer:

    3/4 → Option D.
  5. Quick Check:

    Box A is more likely to produce a red (3/5 vs 1/5), so after seeing red the posterior favors A → 3/4 ✅
Hint: Use Bayes: posterior ∝ prior × likelihood, then normalize.
Common Mistakes: Forgetting to weight by the prior probability of choosing each box.

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