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Independent and Dependent Events

Introduction

In probability, events can be independent or dependent. Knowing which type you are dealing with is crucial because it determines whether probabilities multiply directly or must be adjusted based on prior outcomes.

This pattern helps you understand how to handle multiple events occurring together - like tossing coins, drawing cards, or selecting objects with or without replacement.

Pattern: Independent and Dependent Events

Pattern

Independent events do not affect each other's outcome, while dependent events do.

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) Two coins are tossed together. What is the probability of getting two heads?
(ii) Two cards are drawn from a deck without replacement. What is the probability that both are Kings?

Solution

  1. Part (i) - Understand independence before calculating

    Step 1: Identify the event type
    Each coin toss is independent. Probability of Head on one toss = 1/2.

    Step 2: Multiply probabilities for independent events
    Multiply probabilities of each independent event: (1/2) × (1/2) = 1/4.

    Final Answer: Probability of two heads = 1/4.

    Quick Check:
    HH is 1 out of 4 total outcomes → 1/4 ✅

  2. Part (ii) - Understand dependence before calculating

    Step 1: Compute probability of the first event
    Total cards = 52, Kings = 4. So, P(1st King) = 4/52 = 1/13.

    Step 2: Update probability after card removal
    After removing one King, 3 remain from 51 cards → P(2nd King | 1st King) = 3/51.

    Step 3: Multiply probabilities for dependent events
    Multiply the dependent probabilities: (1/13) × (3/51) = 3 / 663 = 1/221.

    Final Answer: Probability both are Kings = 1/221.

    Quick Check:
    Matches earlier combination approach for two Kings → 1/221 ✅

Quick Variations

1. Tossing multiple coins → independent events.

2. Drawing cards without replacement → dependent events.

3. Picking coloured balls from a bag with replacement → independent.

4. Picking without replacement → dependent.

Trick to Always Use

  • Step 1: Check if one event affects another - if yes → dependent.
  • Step 2: For independent → multiply direct probabilities.
  • Step 3: For dependent → multiply with adjusted conditional probability (P(B|A)).

Summary

Summary

In the Independent and Dependent Events pattern:

  • Independent → events don’t influence each other: P(A ∩ B) = P(A) × P(B).
  • Dependent → first event affects second: P(A ∩ B) = P(A) × P(B | A).
  • With replacement = independent; Without replacement = dependent.
  • Always analyze before applying the correct formula.

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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