Welcome to our article, where simple probability formulas with sample questions.

**1. Probability** deals with the measure or estimation of events that are likely to happen. In mathematics, probability is the calculation of uncertainty.

\( probability=\displaystyle\frac{Number \ of \ favorable \ outcomes}{Total \ numbers \ of \ outcomes} \)

● The probability of an event can only be between 0 and 1 and can also be written as a percentage.

● The probability of event A is often written as P(A).

● If P(A) > P(B), then event A has a higher chance of occurring than event B.

● If P(A) = P(B), then events A and B are equally likely to occur.

**Exercise**: What is the probability of getting head when tossing a coin?

**Answer**: Sample Space = {H, T}

Number of possible outcomes = 2

Number of favorable outcomes = 1(because of only one head).

\( probability=\displaystyle\frac{Number \ of \ favorable \ outcomes}{Total \ numbers \ of \ outcomes} \)

Probability of getting head is \( \displaystyle\frac{1}{2} \)

**2. Addition Rule 1:** When two events, A and B, are mutually exclusive, the probability that A or B will occur is the sum of the probability of each event.

P(A or B) = P(A) + P(B)

**Exercise**: A single 6-sided dice is rolled. What is the probability of rolling a 2 or a 5?

**Answer**: Probability of getting 2, P(2)= 1/6

Probability of getting 6, P(6)= 1/6

Probability of getting 2 or 6,

P(2 or 6)=P(2)+P(6)

\( =\displaystyle\frac{1}{6}+\frac{1}{6}=\frac{2}{6}⇒\frac{1}{3} \)

**3. Addition Rule 2:** When two events, A and B, are non-mutually exclusive, the probability that A or B will occur is:

P(A or B) = P(A) + P(B) – P(A and B)

**Exercise**: In a math class of 30 students, 17 are boys and 13 are girls. On a unit test, 4 boys and 5 girls made an A grade. If a student is chosen at random from the class, what is the probability of choosing a girl or an A student?

**Answer**: Probabilities:

P(girl or A)

= P(girl) + P(A) – P(girl and A)

\( \displaystyle\frac{13}{30}+\frac{9}{30}-\frac{5}{30}=\frac{17}{30} \)

**4. Multiplication Rule 1:**

If A and B are two independent events in a probability experiment, then the probability that both events occur simultaneously is:

P(A and B)=P(A).P(B)

**Exercise**: You have a cowboy hat, a top hat, and an Indonesian hat called a songkok. You also have four shirts: white, black, green, and pink. If you choose one hat and one shirt at random, what is the probability that you choose the songkok and the black shirt?

**Answer**: The two events are independent events; the choice of hat has no effect on the choice of shirt. There are three different hats, so the probability of choosing the songkok is 1/3

There are four different shirts, so the probability of choosing the black shirt is 1/4 . So, by the Multiplication Rule:

P(songok and black shirt)= \( \displaystyle\frac{1}{3}.\frac{1}{4}=\frac{1}{12} \)

**5. Multiplication Rule 2:**

If A and B are two dependent events in a probability experiment, then the probability that both events occur simultaneously is:

P(A and B)=P(A).P(B | A)

(The notation P(B | A) means “the probability of B, given that A has happened.”)

**Exercise**: A bag has 4 white cards and 5 blue cards. We draw two cards from the bag one by one without replacement. Find the probability of getting both cards white.

**Answer**: Let A = event that first card is white and B = event that second card is white.

From question, P(A) = 4/9

Now P (B) = P(B|A) because the events given are dependent on each other.

P (B) = 3/8

So, P(A and B) = P(A) × P(B|A)

= \( \displaystyle\frac{4}{9}x\frac{3}{8}=\frac{1}{6} \)

**6. Bayes’ Theorem:**

Bayes’ theorem is stated mathematically as the following equation:

\( P(A|B)=\displaystyle\frac{P(B|A)P(A)}{P(B)} \)

where A and B are events and P(B) ≠ 0.

● P(A | B) is a conditional probability: the likelihood of event A occuring that B is true.

●P(B | A) is also a conditional probability: the likelihood of event B occuring that A is true.

●P(A) and P(B) are probabilities of observing A and B independently of each other; this is known as marginal probability.

**Exercise**: If dangerous fires are rare (2%) but smoke is fairly common (20%) due to barbecues, and 80% of dangerous fires make smoke. Then what is the Probability of dangerous Fire when there is Smoke?

**Answer**: \( P(Fire|Smoke)=\displaystyle\frac{P(Fire|Smoke)P(Fire)}{P(Smoke)} \)

\( \displaystyle\frac{2\%x80\%}{20\%}=8\% \)

So the “Probability of dangerous Fire when there is Smoke” is 8%