GCSE Maths Practice: probability-scale

Understanding Multi-Step Probability with Changing Sample Spaces

This question explores a common higher-level GCSE probability topic: calculating the chance of two dependent events both occurring when drawing from a finite sample space. Because cards are drawn without replacement, the total number of available cards changes after the first card is drawn. This means each probability must be calculated carefully, keeping track of the reduced sample space and remaining favourable outcomes.

In a standard deck, there are 26 red cards (hearts and diamonds) and 26 black cards (clubs and spades). To find the probability that two cards drawn consecutively are the same colour, we need to consider both scenarios that satisfy this requirement: drawing two red cards or drawing two black cards. These situations are mutually exclusive, meaning they cannot occur at the same time, so their probabilities can be added together once calculated.

Step-by-Step Method

1. Calculate the probability of drawing two red cards. The first draw has 26 favourable outcomes out of 52. Removing that red card leaves 25 red cards out of 51 total cards. Multiply these probabilities: \((26/52)(25/51)\).
2. Do the same for two black cards. The reasoning is identical because the number of black cards is the same as the number of red cards. So the probability is also \((26/52)(25/51)\).
3. Add the two probabilities. Since there are two valid paths to the same final outcome, we sum them: \(2(26/52)(25/51)\).

Once simplified, this produces \(25/51\), showing that the probability of drawing two cards of the same colour is slightly less than 50%. This may feel counterintuitive at first, but it reflects how evenly balanced the deck is and how the second draw's restricted sample space influences the outcome.

Worked Example 1

If we instead wanted the probability of drawing two picture cards (Jack, Queen, or King), the method would be similar. There are 12 picture cards. First draw: \(12/52\). The second draw, without replacement, becomes \(11/51\). Multiply: \((12/52)(11/51)\). Only one scenario exists, so no doubling is needed.

Worked Example 2

Imagine drawing two number cards that are both even. There are 20 even-value cards in a deck. The probability becomes \((20/52)(19/51)\). Again, since there is only one type of event, we do not double the probability.

Common Mistakes

• Forgetting to reduce the total number of cards from 52 to 51 after the first draw.
• Incorrectly assuming that the probability of red–red equals black–black by chance, rather than recognising it is due to equal counts.
• Multiplying by 2 too early or too late, leading to an incorrect fraction.
• Adding probabilities for events that are not mutually exclusive.

Real-Life Applications

Problems like this model real situations where items are taken from a set without replacement: selecting samples in quality control, distributing coloured items in manufacturing, choosing players for teams without repetition, or interpreting probabilities in card-based games. Understanding how to adjust probabilities as the sample space changes is fundamental to probability-based reasoning beyond the classroom.

FAQ

Q: Why do red–red and black–black have the same probability?
A: Because the numbers of red and black cards are identical (26 each).

Q: Why multiply the two probabilities?
A: Because both events must occur in sequence—this is an 'and' event, so multiplication is required.

Q: Why add the two probabilities at the end?
A: Because there are two separate ways to achieve the outcome, and they cannot happen simultaneously.

Study Tip

When solving any multi-step probability question, always list each valid pathway separately, calculate them individually, and combine them only if appropriate. This approach reduces errors and helps you visualise the structure of the problem clearly.