Best Way to Learn Probability

Introduction best way to learn probability:

The theory of best way of probability begins to develop for study of games of chance such as roulette and cards. The best way of probability used for not only games, probability also prevails in other walks of life such as commerce, financial system, and even in day-to-day daily activities. Systematic method for probability theory was introduced by French mathematicians Blaise Pascal and Pierre. In this article we shall discuss best way to learn probability with example problems.


Learn probability formula with example problems


Learn probability formula:

Probability P(E) =        Number of way the event happen
The total number of possible outcome of an event

Example 1: A spinner has 5 equal sectors colored blue, white, green, orange and red. Spinning the spinner, Find the probability of leading green color?

Solution:

The possible outcomes of these events are blue, white, green, orange and red.

P(green) = Number of way land green
Total number of colors

= `1/5.`

Example 2: A six sided unbiased die is rolled. Find the probability getting one? Find the probability of an even number rolling?

Solution:

P(E)  = Number of way to get a one
Total number of sides

P(getting number 1)    = 1.

Total number of sides = 6

Therefore probability   = `1/6.`

Example 3: A six sided unbiased die is rolled. Find the probability of an even number rolling?

Solution:

P(E)= Number of way to get even number
Total number of sides

P (getting even number) = 3.

Total number of sides    = 6

Therefore probability      = 3/6 = 1/2

Example 4: If a1 and a2 are two events related with a random experiment such that P(a2)=0.45, P(a1 or a2)= 0.75 and P(a1 and a2)=0.25, Calculate P(a1).

Solution:

Let P (a1) = x then,

P (a1 or a2) =P (a) + P (a2) – P (a1 and a2)

= 0.75 = x + 0.45 – 0.25

Sum the experiment value

x = (0.75-0.45 + 0.25) = 0.65

Hence P(a1) = 0.55.

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Learn probability with practice problem


Problem 1: A six sided unbiased die is rolled. Find the probability of an odd number rolling?

Answer: `1/2`

Problem 2: A spinner has 4 equal sectors colored blue, white, green, and orange. Spinning the spinner, find the probability of leading white color?

Answer: `1/4.`

Continuous Probability Learning

Introduction to continuous probability learning :

Definition: Learning to define continuous probability of an event occurs when the one or more events occurred. Consider any two probabilities A and B. when the event A occur and it depends on the other event which is already occurred then we can say the conditional probability of above two events are P(A | B). Learning  conditional probability of events using the following formula :

`P((A)/(B))=(P(A U B))/(P(A))`

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Learning to solve conditional probability problems 1:


Learning some problems to find the conditional probability.

Pro 1:Consider a population, the probability a men life at least 70 years is 0.70 and is 0.65.if he won’t live more than 80 years. If a man is 70 years old, find the conditional probability that he will survive on 80 years. If A subset of B then P (A U B) = P (A)

Solution:Let us take A is the event that he lives to 70 years and B is the event that he will live at least 80 years.

So given that P (B) = 0.55 and P (A) = 0.70

So Conditional probability P (`(B)/(A)) = ( P (A U B)) / ( P (A))`

The given condition is P (A and B) = P (B) = 0.65

Conditional probability P (`(A)/(B)` ) = `(0.65)/(0.70)`

P (`(A)/(B)` ) = 0.9286

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Learning to solve conditional probability problems 2:

A box contains red and blue marbles. We are choosing two marbles without replacement. Probability of choosing red and blue marbles is 0.45 and choosing the red marbles on the first draw is 0.57. Find the probability if the second marble is blue if the first one is red?

Solution:Probability of choosing red and blue marbles is 0.45

Probability of choosing red in the first draw is 0.57

So probability of choosing second marble ids blue then the probability is

P (`(Blue)/(Red) = ( P ( Red and Blue))/ (P(Red))`

P (`(B)/(R)` ) = `(0.45)/(0.57)`

P (`(B)/(R)` ) = 0.79 = 79 %