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Independent and Dependent Events

Introduction

Probability में events या तो independent होते हैं या dependent। यह पहचानना बहुत ज़रूरी है क्योंकि इससे तय होता है कि probabilities सीधे multiply होंगी या फिर पहले outcome के आधार पर adjust करनी होंगी।

यह pattern आपको समझने में मदद करता है कि कई events एक साथ होने पर कैसे handle किए जाते हैं - जैसे coins toss करना, cards निकालना, या objects को replacement के साथ या बिना चुनना।

Pattern: Independent and Dependent Events

Pattern

Independent events एक-दूसरे के outcome को affect नहीं करते, जबकि dependent events में एक event दूसरे को affect करता है।

Formulas:
Independent Events: P(A ∩ B) = P(A) × P(B)
Dependent Events: P(A ∩ B) = P(A) × P(B | A)

Step-by-Step Example

Question

(i) दो coins एक साथ toss किए जाते हैं। Two Heads मिलने की probability क्या है?
(ii) 52 cards की deck से बिना replacement दो cards निकाले जाते हैं। दोनों Kings होने की probability क्या है?

Solution

  1. Part (i) - Calculate करने से पहले independence समझें

    Step 1: Event type पहचानें
    हर coin toss independent होता है। Head आने की probability = 1/2.

    Step 2: Independent events की probabilities multiply करें
    (1/2) × (1/2) = 1/4.

    Final Answer: Two Heads की probability = 1/4.

    Quick Check:
    HH एक outcome है total 4 outcomes में से → 1/4 ✅

  2. Part (ii) - Calculate करने से पहले dependence समझें

    Step 1: पहले event की probability निकालें
    Total cards = 52, Kings = 4 → P(1st King) = 4/52 = 1/13.

    Step 2: Card हटाने के बाद probability update करें
    एक King निकलने के बाद 3 Kings बचते हैं और total cards = 51 → P(2nd King | 1st King) = 3/51.

    Step 3: Dependent events की probabilities multiply करें
    (1/13) × (3/51) = 3/663 = 1/221.

    Final Answer: Probability कि दोनों Kings हों = 1/221.

    Quick Check:
    Two Kings वाले combination result से match करता है → 1/221 ✅

Quick Variations

1. कई coins toss करना → independent events।

2. Cards बिना replacement निकालना → dependent events।

3. Bag से coloured balls replacement के साथ निकालना → independent।

4. Replacement बिना निकालना → dependent।

Trick to Always Use

  • Step 1: देखें क्या एक event दूसरे को affect कर रहा है - अगर हाँ → dependent।
  • Step 2: Independent events में probabilities सीधे multiply करें।
  • Step 3: Dependent events में conditional probability (P(B|A)) के साथ multiply करें।

Summary

Summary

Independent and Dependent Events pattern में:

  • Independent → events एक-दूसरे को affect नहीं करते: P(A ∩ B) = P(A) × P(B).
  • Dependent → पहला event दूसरे को affect करता है: P(A ∩ B) = P(A) × P(B | A).
  • With replacement = independent; Without replacement = dependent.
  • हमेशा पहले event type पहचानें, फिर सही formula apply करें।

Practice

(1/5)
1. Two coins are tossed simultaneously. What is the probability that both show heads?
easy
A. 1/2
B. 1/3
C. 1/4
D. 1/6

Solution

  1. Step 1: Identify total outcomes

    For two coins → total ordered outcomes = {HH, HT, TH, TT} = 4.

  2. Step 2: Identify favourable outcomes

    Favourable outcome for both heads = {HH} → 1.

  3. Step 3: Compute probability

    P(both heads) = favourable ÷ total = 1 ÷ 4 = 1/4.

  4. Final Answer:

    1/4 → Option C.

  5. Quick Check:

    Independent events: (1/2) × (1/2) = 1/4 ✅
Hint: Multiply probabilities of independent events (1/2 × 1/2 for coin tosses).
Common Mistakes: Assuming outcomes are not equally likely or forgetting to multiply probabilities.
2. A coin is tossed twice. What is the probability that the first toss is a Head and the second is a Tail?
easy
A. 1/4
B. 1/2
C. 1/3
D. 1/8

Solution

  1. Step 1: Recognise independence

    Each toss is independent; probabilities don't change between tosses.

  2. Step 2: Individual probabilities

    P(Head) = 1/2, P(Tail) = 1/2.

  3. Step 3: Multiply

    P(Head then Tail) = (1/2) × (1/2) = 1/4.

  4. Final Answer:

    1/4 → Option A.

  5. Quick Check:

    Out of {HH, HT, TH, TT}, HT is one → 1/4 ✅
Hint: For ordered independent events, multiply single-event probabilities.
Common Mistakes: Treating independent events as dependent and changing probabilities incorrectly.
3. A card is drawn from a deck, replaced, and another card is drawn. What is the probability that both are Aces?
easy
A. 1/169
B. 1/221
C. 1/13
D. 1/26

Solution

  1. Step 1: Recognise independence (replacement)

    Replacement makes the two draws independent; probabilities stay the same.

  2. Step 2: Single-draw probability

    P(Ace on one draw) = 4/52 = 1/13.

  3. Step 3: Multiply

    P(both Aces) = (1/13) × (1/13) = 1/169.

  4. Final Answer:

    1/169 → Option A.

  5. Quick Check:

    Replacement keeps probability constant → (1/13)² = 1/169 ✅
Hint: With replacement → square the single-draw probability for two identical events.
Common Mistakes: Using 52 for second draw (no replacement) or forgetting independence.
4. Two cards are drawn successively from a deck without replacement. What is the probability that both are Queens?
medium
A. 1/169
B. 1/1326
C. 1/325
D. 1/221

Solution

  1. Step 1: Recognise dependence

    Without replacement → the second draw depends on the first.

  2. Step 2: First draw probability

    P(1st Queen) = 4/52 = 1/13.

  3. Step 3: Conditional second draw

    After drawing one Queen, remaining Queens = 3 and remaining cards = 51 → P(2nd Queen | 1st Queen) = 3/51.

  4. Step 4: Multiply

    P(both Queens) = (1/13) × (3/51) = 3/663 = 1/221.

  5. Final Answer:

    1/221 → Option D.

  6. Quick Check:

    Equivalent to 4C2 ÷ 52C2 = 6/1326 = 1/221 ✅
Hint: For without replacement, compute P(A) × P(B|A) stepwise.
Common Mistakes: Using same denominator for both draws (ignoring reduced deck).
5. A bag contains 5 red and 3 blue balls. Two balls are drawn one after another without replacement. What is the probability that both are red?
medium
A. 1/3
B. 5/14
C. 5/7
D. 10/21

Solution

  1. Step 1: Recognise dependence

    No replacement → second draw probability changes based on first draw.

  2. Step 2: First draw probability

    P(1st red) = 5/8.

  3. Step 3: Conditional second draw

    If first was red, remaining red = 4 and total = 7 → P(2nd red | 1st red) = 4/7.

  4. Step 4: Multiply

    P(both red) = (5/8) × (4/7) = 20/56 = 5/14.

  5. Final Answer:

    5/14 → Option B.

  6. Quick Check:

    Adjust denominator after first draw → 5/8 × 4/7 = 5/14 ✅
Hint: Reduce the total after each draw and multiply conditional probabilities.
Common Mistakes: Treating without replacement as with replacement (keeping denominator fixed).

Mock Test

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