GCSE Maths Practice: probability-scale

Understanding Probability with Sequential Draws

This type of problem is a classic example of conditional probability in GCSE Maths. It tests your ability to handle events that take place one after another without replacement. When objects are removed from a group, the sample space changes, and this affects the probability of all subsequent events. In this question, a bag contains 3 green balls and 5 red balls for a total of 8 balls. The task is to find the probability of drawing a green ball first and then a red ball from what remains.

One of the crucial ideas here is recognising when events are dependent. Two events are dependent when the outcome of the first affects the probabilities of the second. Drawing without replacement is always a dependent scenario because removing an item reduces the total number of possible outcomes. For this reason, you cannot simply multiply two identical fractions such as \(5/8 \times 5/8\). Instead, you must adjust the second probability after updating the number of items in the bag.

Breakdown of the Events

Event 1 is drawing a green ball. Since there are 3 green balls out of 8, the probability is:

\[ P(\text{green first}) = \frac{3}{8}. \]

Once a green ball is removed, only 7 balls remain in the bag. The red balls are unaffected, so there are still 5 red balls left. This makes the probability of drawing a red ball second:

\[ P(\text{red second | green first}) = \frac{5}{7}. \]

Because both events must occur in sequence, and because the second depends on the first, the total probability is the product of the two individual probabilities:

\[ P(\text{green then red}) = \frac{3}{8} \times \frac{5}{7}. \]

Worked Example 1

Suppose the question asked for the probability of drawing two red balls in a row. First, the probability of red is \(5/8\). After removing one red ball, the probability of drawing another red becomes \(4/7\). The total probability is \((5/8)(4/7)\).

Worked Example 2

If instead you wanted the probability of drawing a green ball followed by another green ball, you begin with \(3/8\), then after removing one green ball, the new probability is \(2/7\). The combined probability is \((3/8)(2/7)\).

Common Mistakes

• Not reducing the total number of balls after the first draw.
• Accidentally reducing the number of red balls even though a green ball was removed.
• Using \(3/8\) again for the second draw instead of adjusting to \(3/7\) or \(5/7\).
• Adding probabilities instead of multiplying them.

Real-Life Applications

Dependent probabilities appear in many real-world scenarios: selecting items without replacement in quality testing, choosing cards from a deck, sampling biological populations, and even drawing raffle tickets. Understanding how to adjust the sample space after each step is essential in statistics, data analysis, and many practical decision-making situations.

FAQ

Q: Why multiply the probabilities?
Because both events must happen in order. In probability, “and” means multiply.

Q: Why does the denominator change to 7?
Because one ball is removed and not returned.

Q: Do we ever add probabilities?
Only when events are mutually exclusive (cannot both happen). This is not one of those cases.

Study Tip

Whenever you see “without replacement,” immediately rewrite the new totals for the next event. This avoids the most common errors and ensures your calculations follow the correct probability rules.