krit.club logo

Statistics and Probability - Probability of Single and Independent Events

Grade 8IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Probability is the numerical measure of the likelihood that an event will occur, ranging from 0 (impossible) to 1 (certain).

•

The Sum of Probabilities: The sum of the probabilities of all possible mutually exclusive outcomes in an experiment is always equal to 1.

•

Complementary Events: The probability that an event will not occur is 1 minus the probability that it will occur.

•

Mutually Exclusive Events: Two events that cannot happen at the same time. If A and B are mutually exclusive, P(A or B) = P(A) + P(B).

•

Independent Events: Two events are independent if the outcome of the first event does not affect the outcome of the second event.

•

Sample Space: A list or set of all possible outcomes of a probability experiment.

•

Relative Frequency: An estimate of probability based on experimental data (Total successes / Total trials).

📐Formulae

P(E)=Number of successful outcomesTotal number of possible outcomesP(E) = \frac{\text{Number of successful outcomes}}{\text{Total number of possible outcomes}}

P(A′)=1−P(A)P(A') = 1 - P(A)

P(A∩B)=P(A)×P(B)P(A \cap B) = P(A) \times P(B) (Multiplication rule for independent events)

P(A∪B)=P(A)+P(B)P(A \cup B) = P(A) + P(B) (Addition rule for mutually exclusive events)

💡Examples

Problem 1:

A bag contains 4 red, 3 blue, and 5 yellow sweets. One sweet is chosen at random. What is the probability that it is yellow?

Solution:

512\frac{5}{12}

Explanation:

First, find the total number of outcomes: 4+3+5=124 + 3 + 5 = 12. The number of successful outcomes (yellow) is 5. Using the formula P(E)=n(E)n(S)P(E) = \frac{n(E)}{n(S)}, the probability is 5/125/12.

Problem 2:

The probability that it rains tomorrow is 0.2. What is the probability that it does not rain?

Solution:

0.80.8

Explanation:

This uses the concept of complementary events. P(not rain)=1−P(rain)P(\text{not rain}) = 1 - P(\text{rain}). Therefore, 1−0.2=0.81 - 0.2 = 0.8.

Problem 3:

A fair coin is flipped and a standard 6-sided die is rolled. Find the probability of getting a 'Tail' and a 'Prime Number'.

Solution:

14\frac{1}{4}

Explanation:

These are independent events. P(Tail)=12P(\text{Tail}) = \frac{1}{2}. The prime numbers on a die are 2, 3, and 5, so P(Prime)=36=12P(\text{Prime}) = \frac{3}{6} = \frac{1}{2}. For independent events, multiply the probabilities: 12×12=14\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}.

Problem 4:

Two independent events, A and B, have probabilities P(A)=0.4P(A) = 0.4 and P(B)=0.7P(B) = 0.7. Find P(A and B)P(A \text{ and } B).

Solution:

0.280.28

Explanation:

Since the events are independent, we use the multiplication rule: P(A and B)=P(A)×P(B)=0.4×0.7=0.28P(A \text{ and } B) = P(A) \times P(B) = 0.4 \times 0.7 = 0.28.