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CIE IGCSE Maths Study Notes

8.3.2 Combined Events without Replacement

Understanding how probabilities change in scenarios where events occur without replacement is crucial for mastering the topic of probability. This section delves into the fundamental principles and calculations that govern these situations, illustrating the impact of each event on the sample space.

Combined Events without Replacement Illustration

Image courtesy of Third Space learning

Introduction

When we discuss combined events without replacement, we're exploring scenarios where the outcome of one event affects the outcomes of subsequent events. This adjustment to the sample space is key to calculating probabilities accurately in such contexts.

Key Concepts

  • Sample Space Reduction: After each event, the total number of outcomes in the sample space decreases.
  • Probability Adjustment: The probabilities of subsequent events need adjustment based on the outcomes of previous events.

Worked Examples

Example 1: Drawing Cards from a Deck

You are drawing cards from a standard deck of 52 without replacement. What is the probability of drawing a heart followed by a spade?

Deck of Cards

Solution:

  • Total initial cards: 52
  • Cards that are hearts: 13
  • Probability of first draw being a heart: P(Heart)=1352P(\text{Heart}) = \dfrac{13}{52}

After drawing a heart, the sample space reduces:

  • Total cards now: 51
  • Cards that are spades: 13
  • Probability of second draw being a spade: P(SpadeHeart)=1351P(\text{Spade}|\text{Heart}) = \dfrac{13}{51}

Combined probability: P(Heart then Spade)=P(\text{Heart then Spade}) = P(Heart)×P(SpadeHeart)=1352×1351P(\text{Heart}) \times P(\text{Spade}|\text{Heart}) = \dfrac{13}{52} \times \frac{13}{51} =

Example 2: Removing Balls from a Bag

A bag contains 5 red balls and 3 blue balls. If two balls are removed successively without replacement, what is the probability that both balls are red?

Bag of Red and Blue Balls

Solution:

  • Total balls initially: 8
  • Probability of first ball being red: P(Red1)=58P(\text{Red}_1) = \dfrac{5}{8}
  • After removing a red ball:
  • Total balls now: 7
  • Probability of second ball being red: P(Red2Red1)=47P(\text{Red}_2|\text{Red}_1) = \dfrac{4}{7}

Combined probability:

P(Red1 and Red2)=P(Red1)×P(Red2Red1)=58×47P(\text{Red}_1 \text{ and } \text{Red}_2) = P(\text{Red}_1) \times P(\text{Red}_2|\text{Red}_1) = \frac{5}{8} \times \frac{4}{7}

Calculating Probabilities

To calculate these probabilities effectively, one must:

1. Determine the total number of possible outcomes for the initial event.

2. Update the sample space based on the outcome of the first event.

3. Calculate the probability of the subsequent event based on the updated sample space.

4. Multiply the probabilities of the individual events to find the combined probability.

Practice Questions

Question 1

A box contains 6 chocolates, 4 of which are dark chocolate and 2 are milk chocolate. If two chocolates are taken at random without replacement, what is the probability that both are dark chocolate?

Chocolates Illustration

Solution:

  • Calculate the probability of the first chocolate being dark.
  • Update the sample space and calculate the probability of the second chocolate being dark.
  • Multiply both probabilities for the final answer.

Question 2

In a lottery draw, 100 tickets are sold, and 10 of these tickets are winners. If two tickets are drawn one after the other without replacement, what is the probability that both tickets are winners?

100 Tickets

Solution:

  • Determine the probability of the first ticket being a winner.
  • Update the sample space for the second draw.
  • Calculate the probability of the second ticket being a winner.
  • Multiply the probabilities for the final answer.

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