How to Solve: Probability & Probability Distributions (Binomial, Normal, Conditional Probability)

How to Solve: Probability & Probability Distributions (Binomial, Normal, Conditional Probability)

GCSE / A-Level Maths

Introduction

"Mastering probability doesn’t just help you pass your exam—it lets you predict real-world outcomes, from medical test accuracy to sports betting odds. On your GCSE/A-Level paper, probability questions can be worth 10-15% of your total marks—so one mistake could cost you a grade. Today, we’ll break it down into simple, repeatable steps so you can solve any question with confidence."

What You Need To Know First

Before diving in, make sure you understand: 1. Basic probability rules (e.g., P(A) = number of favourable outcomes / total outcomes). 2. Tree diagrams and Venn diagrams (for visualising events). 3. Percentages and decimals (probabilities are often given as decimals between 0 and 1).

Key Vocabulary

Formulas To Know

1. Basic Probability

Formula: [ P(A) = \frac{\text{Number of ways A can happen}}{\text{Total possible outcomes}} ]

Variables: - ( P(A) ) = Probability of event A happening.

MEMORISE THIS

2. Independent Events (AND Rule)

Formula: [ P(A \text{ and } B) = P(A) \times P(B) ]

Variables: - ( P(A) ) = Probability of event A. - ( P(B) ) = Probability of event B.

MEMORISE THIS

3. Mutually Exclusive Events (OR Rule)

Formula: [ P(A \text{ or } B) = P(A) + P(B) ]

Variables: - ( P(A) ) = Probability of event A. - ( P(B) ) = Probability of event B.

MEMORISE THIS

4. Conditional Probability

Formula: [ P(A|B) = \frac{P(A \text{ and } B)}{P(B)} ]

Variables: - ( P(A|B) ) = Probability of A given B has happened. - ( P(A \text{ and } B) ) = Probability of both A and B happening. - ( P(B) ) = Probability of B happening.

MEMORISE THIS

5. Binomial Probability

Formula: [ P(X = r) = \binom{n}{r} p^r (1-p)^{n-r} ]

Variables: - ( n ) = Number of trials. - ( r ) = Number of successes. - ( p ) = Probability of success on one trial. - ( \binom{n}{r} ) = "n choose r" (number of ways to choose r successes from n trials).

GIVEN ON EXAM SHEET (but know how to use it!)

6. Normal Distribution

Formula: - Standardising (Z-score): [ Z = \frac{X - \mu}{\sigma} ] - Probability from tables: Use the Z-table to find ( P(Z \leq z) ).

Variables: - ( X ) = Value from the dataset. - ( \mu ) = Mean of the distribution. - ( \sigma ) = Standard deviation.

GIVEN ON EXAM SHEET (but practice using it!)

Step-by-Step Method

How to Solve ANY Probability Question

1. Read the question carefully. Underline key numbers and what’s being asked.
2. Identify the type of probability:
3. Basic? (Single event)
4. Independent/dependent? (AND/OR rules)
5. Conditional? (Given that…)
6. Binomial? (Fixed trials, success/failure)
7. Normal? (Bell curve, mean, standard deviation)
8. Write down what you know. List all given probabilities, ( n ), ( p ), ( \mu ), ( \sigma ), etc.
9. Choose the right formula. Match the question to one of the formulas above.
10. Plug in the numbers. Substitute values into the formula.
11. Calculate carefully. Use a calculator for binomial coefficients or Z-scores.
12. Check your answer. Does it make sense? (Probabilities must be between 0 and 1.)
13. Write a clear conclusion. Answer the question in words (e.g., "The probability is 0.35").

Worked Example Using the Steps

Question: A bag contains 5 red balls and 3 blue balls. Two balls are drawn without replacement. What is the probability that both are red?

Step-by-Step Solution:

1. Read the question. Underline:
2. 5 red, 3 blue balls.
3. Two balls drawn without replacement.

4. Find probability both are red.

5. Identify the type:

6. Dependent events (probability changes after first draw).

7. Use AND rule for dependent events:
   [ P(\text{Red and Red}) = P(\text{First Red}) \times P(\text{Second Red | First Red}) ]

8. Write down what you know:

9. Total balls = 5 red + 3 blue = 8.
10. ( P(\text{First Red}) = \frac{5}{8} ).
11. After first red is drawn, 4 red and 3 blue left.

12. ( P(\text{Second Red | First Red}) = \frac{4}{7} ).

13. Choose the right formula:

14. ( P(\text{Red and Red}) = P(\text{First Red}) \times P(\text{Second Red | First Red}) ).

15. Plug in the numbers:
    [ P(\text{Red and Red}) = \frac{5}{8} \times \frac{4}{7} ]

16. Calculate:
    [ \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14} ]

17. Check:

18. Probability is between 0 and 1.

19. Makes sense (less than 50% since blue balls are present).

20. Conclusion:

21. The probability that both balls are red is ( \frac{5}{14} ).

Worked Examples

Example 1 – Basic (Independent Events)

Question: A fair coin is flipped and a fair die is rolled. What is the probability of getting a head and a 4?

Solution: 1. Identify type: Independent events (coin and die don’t affect each other). 2. Use AND rule for independent events:
[ P(\text{Head and 4}) = P(\text{Head}) \times P(4) ] 3. Calculate:
- ( P(\text{Head}) = \frac{1}{2} )
- ( P(4) = \frac{1}{6} )
- ( \frac{1}{2} \times \frac{1}{6} = \frac{1}{12} )

Answer: ( \frac{1}{12} )

What we did and why: - Recognised independent events → used multiplication rule. - Simple probabilities → no need for conditional or binomial.

Example 2 – Medium (Conditional Probability)

Question: In a class, 60% of students are girls. 25% of girls and 10% of boys have brown hair. If a student is chosen at random and has brown hair, what is the probability they are a girl?

Solution: 1. Identify type: Conditional probability (given brown hair, find probability of girl). 2. Use formula:
[ P(\text{Girl | Brown Hair}) = \frac{P(\text{Girl and Brown Hair})}{P(\text{Brown Hair})} ] 3. Calculate ( P(\text{Girl and Brown Hair}) ):
- ( P(\text{Girl}) = 0.6 )
- ( P(\text{Brown Hair | Girl}) = 0.25 )
- ( 0.6 \times 0.25 = 0.15 ) 4. Calculate ( P(\text{Brown Hair}) ):
- ( P(\text{Boy}) = 0.4 )
- ( P(\text{Brown Hair | Boy}) = 0.1 )
- ( P(\text{Boy and Brown Hair}) = 0.4 \times 0.1 = 0.04 )
- ( P(\text{Brown Hair}) = 0.15 + 0.04 = 0.19 ) 5. Plug into formula:
[ \frac{0.15}{0.19} \approx 0.789 ]

Answer: ( 0.789 ) (or ( \frac{15}{19} ))

What we did and why: - Used conditional probability formula. - Calculated total probability of brown hair by considering both girls and boys.

Example 3 – Exam-Style (Binomial Distribution)

Question: A biased coin has a 0.6 probability of landing heads. The coin is flipped 8 times. What is the probability of getting exactly 5 heads?

Solution: 1. Identify type: Binomial distribution (fixed trials, success/failure). 2. Write down values:
- ( n = 8 ) (trials)
- ( r = 5 ) (successes)
- ( p = 0.6 ) (probability of success) 3. Use binomial formula:
[ P(X = 5) = \binom{8}{5} (0.6)^5 (0.4)^3 ] 4. Calculate ( \binom{8}{5} ):
- ( \binom{8}{5} = \binom{8}{3} = 56 ) (given on formula sheet) 5. Calculate probabilities:
- ( (0.6)^5 = 0.07776 )
- ( (0.4)^3 = 0.064 ) 6. Multiply:
[ 56 \times 0.07776 \times 0.064 \approx 0.2787 ]

Answer: ( 0.279 ) (3 d.p.)

What we did and why: - Recognised binomial scenario. - Used the formula correctly, including "n choose r." - Rounded to 3 decimal places (common in exams).

Common Mistakes

Exam Traps

1-Minute Recap

"Here’s your last-minute checklist for probability questions: 1. Basic probability? Just divide favourable outcomes by total outcomes. 2. Independent events? Multiply probabilities. Dependent? Adjust after each event. 3. Conditional probability? Use ( P(A|B) = \frac{P(A \text{ and } B)}{P(B)} ). 4. Binomial? Use ( \binom{n}{r} p^r (1-p)^{n-r} ). Remember: fixed trials, success/failure. 5. Normal distribution? Standardise to Z-scores and use the table. 6. Always check: Probabilities must be between 0 and 1. If not, you’ve gone wrong. 7. Read the question twice. Underline key words like ‘without replacement,’ ‘at least,’ or ‘given that.’ You’ve got this—go smash that exam!