GCSE Maths Practice: probability-scale

Understanding Sequential Probability Without Replacement

This question explores a key GCSE Maths skill: calculating probability for two dependent events occurring one after another. Because the first card is not replaced before drawing the second, the total number of cards changes, which means the second probability depends on the outcome of the first event. This type of problem teaches students how to track changing sample spaces and apply conditional thinking. These skills later appear in more complex probability topics such as tree diagrams, conditional probability formulas, and combined event reasoning.

A standard deck of 52 playing cards contains 26 red cards and 26 black cards. The red cards belong to two suits: hearts and diamonds, each containing 13 cards. The question asks for the probability of drawing a red card first and then drawing a queen. The probability of the first draw is straightforward: there are 26 red cards out of 52. Once a red card is drawn and removed, the deck now contains only 51 cards. Importantly, there are still four queens in the deck because drawing a red card does not remove a queen unless the first card was the queen of hearts or the queen of diamonds. However, because probabilities must account for all possible red-card outcomes, we treat the probability of drawing a queen next as \(4/51\). Conditional probability accounts for this subtlety automatically.

Step-by-Step Method

1. Identify total red cards in the deck: 26.
2. Calculate probability of drawing a red card first: \(26/52\).
3. Recognise that drawing a red card reduces the deck size to 51.
4. Count the total queens remaining: 4.
5. Calculate the probability of drawing a queen next from the 51-card deck: \(4/51\).
6. Multiply the two probabilities to obtain the final answer.

Worked Example 1

Find the probability of drawing a black card followed by a queen. The structure remains the same. P(black first) = 26/52. There are still four queens left after removing a black card, so P(queen second) = 4/51. The overall probability is also \((26/52) \times (4/51)\).

Worked Example 2

If the question asked for drawing a queen first and then another queen without replacement, the first probability is \(4/52\). After removing one queen, three queens remain from a total of 51 cards. The probability becomes \((4/52) \times (3/51)\). This shows how favourable outcomes can change depending on what type of card is drawn.

Common Mistakes

• Forgetting to change 52 to 51 after the first draw.
• Assuming the number of queens changes when a red card is drawn. It only changes if the first card is specifically a queen.
• Adding probabilities instead of multiplying for sequential events.

Real-Life Applications

The logic of sequential, dependent probability applies to real-life situations such as selecting items for quality inspection, drawing raffle tickets, forming committees from fixed groups, or predicting outcomes in card-based games. Understanding how probabilities change when the sample space changes is crucial in modelling real-world probability.

FAQ

Q: Do I need to consider whether the first red card might be a queen?
A: No — the calculation accounts for all red cards equally. The probability simplifies correctly without separating cases.

Q: Why multiply the probabilities?
A: Because both events must happen in sequence; multiplication is the rule for ‘and’ events.

Q: Could the second probability ever be \(3/51\)?
A: Yes, but only if the first card drawn was a queen. Since the first card could be any red card, the simpler combined probability is used.

Study Tip

Always rewrite the sample size after each draw and check how many favourable outcomes remain. This guarantees accuracy when solving multi-step probability questions.