We will go over some basic probability problems in this post.
The problems will include concepts such as venn diagrams and proportions.
College students problem
The study was conducted in awesome university and 600 students were randomly asked a question. on which subjects they studied. 400 said that they studied mathematics, 309 said that they studied economics. 180 said they studied mathematics and economics. Draw a Venn diagram of the given data and find the following values.
1> number of students studying only mathematics.
2> number of students studying only one of mathematics or economics.
3> number of students studying neither mathematics nor economics.
4> Probability of randomly picking a student that studied both subjects.
5> Probability of picking a student that only studied mathematics.
6> Probability of picking a student that studied economics and other subjects but not mathematics.
This Venn diagram shows the division of all 600 students based on their answers. When working on a Venn diagram, we always start working from the intersection and work our way outwards. We have the given number of 180 students that study both Mathematics and Economics. We can use that number to find the number of students that study only Economics and Mathematics, as given in the diagram. When we add all of the numbers in the circles we find out that it adds to only 529. That means there are 71 students from our sample of 600 that do not study mathematics nor economics. We will solve the questions in order.
1> Only Mathematics
We get the answer of this question from our Venn Diagram as 220.
2> Only Mathematics or Economics
This number can be recognized as the addition of Only mathematics + Only Economics = 220 + 129 = 349.
3> Neither Mathematics nor Economics
We have this number from the Venn diagram. it is 71.
Basic Probability part of the problem
We will start the probability part of the problem from here. This problem only deals with Unconditional Probability. You can find the definitions of these basic terms Basics of Probability Part 1 and Part 2.
4> Probability of picking a student that studied both subjects
This Probability can be calculated using the simple formula
P(both) = Number of students that studied both / total number = 180/600 = 0.3
We can also say that there is a 30% chance that a random student picked will have studied both subjects.
5> Probability of only Mathematics
Since this is unconditional probability we again divide the number of students with only mathematics by the total sample. P(only mathematics) = number of only mathematics students / total number = 220/600 = 0.3667
6> Probability of picking a student that studied economics and other subjects but not mathematics
If we study the statement, then we can frame it as Probability of not studying mathematics.
P(not Mathematics) = 1 – P(mathematics)
This can be simplified by the complementary property of probability.
P (not Mathematics) = 1 – (400/600) = 1 – 0.6667 = 0.3333
The Symptoms Problem
A certain pandemic going through the world population has about 5 different symptoms that doctors can use to determine the severity of the disease.
The probability of displaying the number of symptoms is given by the table below.
find the probabilities of the following:
Probability that a patient shows 3 symptoms, 5 symptoms, less than 4 symptoms, at least 3 symptoms, and no more than 1 symptom.
This problem is considered to be a part of discrete basic probability. we will learn more about different probability types in future problems. Discrete probability has random variables of an event and each event has a specific probability value attached with it. The Total event size has a cumulative probability of 1.
We will do an addition to confirm this fact.
0.025 + 0.356 + 0.418 + 0.112 + 0.060 + 0.029 = 1.000
Now we know for certain that this is a valid discrete probability, since the probabilities all add to 1. This can be stated logically. That a person who has the disease will have to exhibit some of the symptoms with the given likelihood it.
To continue, we start with probability of 3 and 5 symptoms.
The probability of showing an exact number of symptoms is given in the table.
P(X=3) = 0.112 and P(X=5) = 0.029
The terms Less than, Greater than, at least, at most, no more than, no less than and similar; are terms that signify that we are meant to do a CUMULATIVE PROBABILITY calculation.
Probability of X less than 4 can be mathematically stated as P(X<4).
P(X<4) = P(X=0) + P(X=1) + P(X=2) + P(X=3) = 0.025 + 0.356 + 0.418 + 0.112 = 0.911
This means that there is a 91.1% chance that a randomly chosen person will exhibit less than 4 symptoms.
Probability of showing at least 3 symptoms. Note that this statement says “At least 3”. So 3 will be part of the cumulative probability.
P(X>=3) = P(X=3) + P(X=4) + P(X=5) = 0.112 + 0.060 +0.029 = 0.201
This means that there is a 20.1% chance that a randomly chosen person will exhibit at least 3 symptoms.
Probability of no more than 1 symptom
P(X <=1) = P(X=0) + P(X=1) = 0.025 + 0.356 = 0.381
This means there is a 38.1% chance that a randomly chosen person will show no more than 1 symptom.
These were two examples of basic probability. Click here to get more examples.