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Conditional Probability

Introduction

Conditional Probability measures the probability of an event occurring given that another event has already occurred. It is one of the most important concepts in probability and is used in real-world contexts like weather forecasting, quality control, and risk analysis.

The key idea is to focus on a restricted sample space - only those outcomes where the first event has occurred are considered when evaluating the second.

Pattern: Conditional Probability

Pattern

The probability of event B given that event A has occurred is given by:

P(B | A) = P(A ∩ B) / P(A)
This formula helps to find the probability of one event under the condition that another related event is already known to have occurred.

Step-by-Step Example

Question

A card is drawn from a standard deck of 52 cards. What is the probability that it is a King, given that it is a face card?

Solution

  1. Step 1: Identify known information

    Total cards = 52. Face cards are {J, Q, K} from 4 suits → 12 face cards.

  2. Step 2: Identify favourable cases

    Among face cards, Kings = 4 (one per suit).

  3. Step 3: Apply conditional probability formula

    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:

    Out of 12 face cards, 4 are Kings → 4/12 = 1/3 ✅

Quick Variations

1. Probability of drawing a red card given the card is a face card.

2. Probability that a student passed Math given they passed Science.

3. Probability of a defective item given it was selected from a specific machine.

Trick to Always Use

  • Step 1: Narrow down the sample space to where the condition is true.
  • Step 2: Use the formula P(B | A) = P(A ∩ B) / P(A).
  • Step 3: Focus on “given that” - it defines your new denominator.

Summary

Summary

In the Conditional Probability pattern:

  • Formula: P(B | A) = P(A ∩ B) / P(A).
  • The “given” event (A) becomes the new sample space.
  • Used when events are related, not necessarily independent.
  • Always verify that P(A) ≠ 0 (you can’t condition on an impossible event).

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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