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%