Math Examples of Probability of Occurrence of an Event

Probability can be defined as chances of occurrence of an event. Probability always lie between 0 and 1. Only in the ideal cases, it can be zero or one. If Probability of happening of an event is higher, it assures that chances of occurrence of that event is also high.
Probability of an event X is written as P (X).  Complement of an event X is the event (not X) that is in the event of X not occurring.
Probability in this case will be given as P (not X) = 1 – P (X).
Joint probability of two events X and Y can be given as:
P (X ∩ Y).
This type of probability shows that both events X and Y happen simultaneously. This probability is known as Intersection probability. If two events are independent then intersection probability is given as:
P (X and Y) = P (X∩ Y) = P (X) P (Y).
If either X or Y or both occur at same time, then it is known as the Union of the events X and Y and is given mathematically as P (X U Y) and if events X and Y are mutually exclusive then probability of occurrence will be given as:
P (X or Y) = P (X U Y) = P (X) + P (Y).
Law of probability is called addition probability in which probability of X or Y equals to addition of probability of X and probability of Y and minus the probability of X and Y from this addition.
Mathematically,
P(X or Y) = P (X) + P (Y) – P (X ∩Y) where P (X ∩ Y) = 0.
 
If 8 red balls,7 blue balls 6 green balls, find the probabilty of picking one ball that is red or green?
Let P(X) = Probability of drawing the red ball & let P (Y) be Probability of drawing the green ball. Then Total outcomes are 17. So, P (X) = 8 / 17 & P(Y) = 6/17. Therefore P (X or Y) = 8/17 + 6/17 = 14/17.
 

For solving this type of problem, we need to have good knowledge of Addition Theorem of Probability. According to addition theorem, if we have two Mutually Exclusive Events then we can find the Union of all the event by the formula given below:
P(A ∪B) = P(A) +P(B)- P(A∩B)
If we talk about our question we have  number of Students who like Engineering  so the probability will be P(A) = 30 and the number of  students who like medical are  40 so P(B) = 40 and the number of students who like both Engineering and medical  are 12 so P(A∩B) = 12 , our task is to find the number of students in the class which can be find easily by determining P(A ∪B), so by the additional formula,
P(A ∪B) = 40 + 50 – 20,
P(A ∪B) = 70, So the total number of students in the class is 70.

For solving this type of problem, we need to have good knowledge of Addition Theorem of Probability. According to addition theorem, if we have two Mutually Exclusive Events then we can find the Union of the entire event by the formula given below:
P (A ∪B) = P (A) +P (B) - P (A∩B)
If we talk about our question we have number of Students who like Math so probability will be P(A) = 30 and the number of  students who like Science are  40 so P(B) = 40 and the number of students who like both math and science are 12 so P(A∩B) = 12 , our task is to find the number of students in the class that we can find by finding P(A ∪B), so by the additional formula,
P (A ∪B) = 40 + 30 – 12
P (A ∪B) = 58
So the total number of student in the class is 58.

For solving this type of problem, we need to have good knowledge of Addition Theorem of Probability. According to addition theorem, if we have two Mutually Exclusive Events then we can find the Union of the entire event by the formula given below:
P (A ∪B) = P (A) +P (B) - P (A∩B)
Here P (A) is the probability of first event and P (B) is probability of second event , P(A∩B) is the probability  of Intersection of both the event and P(A ∪B) is probability of union of both the event.
If we talk about our question we have number of students who like cricket so probability will be P (A) = 20 and the number of  students who like football are twenty so P(B) = 10 and the number of students in the class is 50 so P(A ∪B) = 50. Now, if we see our formula only one thing is unknown that is P(A∩B), so by putting all the values we can  find the value of P(A∩B).
50 = 20 + 10 - P (A∩B)
Now we will take P (A∩B) to right side so there will be a change in sign,
P (A∩B) = 30 – 50
P (A∩B) = -20
There is no meaning of minus sign here, so the players who play both football and cricket are 20.

Since the bag has lemon flavored candies only,
So the Probabilities that is to be calculated as:
An orange flavored candies = favorable flavored / Total number of flavored,
An orange flavored candies = 0 (no orange flavor) / 1 (a lemon flavor only) = 0,
P (a lemon flavor) = 1 (one lemon flavor) /1 (one lemon flavor) = 1.

The total number of bulbs in a box = 600,
The number of defective bulbs = 12,
Number of non defective bulbs in a box = 600 – 12 = 588,
Thus we have n (S) = 600, n(A) = 588,
So the Probability of non defective bulbs is
P (non defective bulb) = n (A) / n (S) = 588 / 600 = 98 / 100 = 0.98.

As we know that a leap year has fifty two complete weeks and two days extra. The two days extra may have the Sample Space
S = S – M, M – T, T – W, W – T, T – F, F – S, S – S,
n (S) = 7,
A = S – M, S – S,
n (A) = 2,
thus,
P(A) = P (a leap year has fifty three Sundays) = n (A) / n (S) = 2 / 7.

Suppose that the number of blue balls are ‘x’.
The Sample Space for the event is given by
S = 5 red balls, x blue balls = n(S) = 5 + x,
x (red balls) = 5 and n (blue balls) = x,
With the help of given condition, we get,
P (a blue ball) = 2 * P (a red ball),
n (a blue ball) / n (S) = 2 * n (a red ball) / n(S),
x / (5 + x) = 2 * 5 / (5 + x),
After solving this equation with the help of cross multiplication we get
x = 10,
Hence the number of blue balls in the container is 10.

This is the simplest problem as we know that,
P (winning a game) + P (not winning a game) = 1,
0.3 + P (not winning a game) = 1,
Thus, P (not winning a game or losing a game) = 1 – 0.3 = 0.7.

Here the Sample Space S = one letter written Cube = n(S) = 6.
P (of setting a cube having letter A) = n(A) / n(S) = 2 / 6 = 1 / 3
Here n(S) = 16 and A = a cube with the letter D that is n (A) = 1
Thus, P (a cube having a letter D) = n (A) / n (S) = 1 / 6.

Here n (S) = 1000 that is n (A) = 5,
P (winning a prize) = n (A) / n(S) = 5 / 1000 = 1 / 200.

The total number of events are thirty six that is n(S) = 36,
Favorable events that is a batsman hits a boundary n (A) = 6,
Thus P(A) = P (a batsman hits a boundary) n(A) / n(S) = 6 / 36 = 1/ 6,
Here P (not A) = P (a batsman did not hit) – 1 – 1 / 6 = 5 / 6.

The total Numbers of events are:
n (S) = 6,
Favorable events = a number less than 6 = A = 1, 2, 3, 4, 5
n (A) = 5,
Thus P (A) = n(A) / n(S) = 5 /6.

The Sample Space of the event is
S = 1, 2, 3, 4, 5, 6
There are six equally likely outcomes, probability of obtaining an ace
= favorable outcomes / Total number of outcomes = 1/ 6.
All the six events are favorable, so the Probability of coming up with a number less than 8 turns up
 = 6 / 6 = 1
Suppose A = an odd number = 1, 3, 5.
Thus n (A) = 3,
The probability of getting an odd number n (A) / n(S) = 3 / 6 = 1 / 6.