# Probability Formulas

## Formulas For Probability

In Aptitude, Probability is a very Important Topic for the Competitive Exam make sure that you will get all the related Probability Formulas with a clear understanding.

On this Page, the Probability Formulas is given for your convenience, so that you can solve the Probability-based Question without any problem.

### Formula & Definition for Probability

• Probability is a number that reflects the chance or possibility of a particular event will occur.
• Probability refers to the extent of occurrence of events. When an event occurs like throwing a ball, picking a card from deck, etc ., then the must be some probability associated with that event.
• In terms of mathematics, probability refers to the ratio of wanted outcomes to the total number of possible outcomes. There are three approaches to the theory of probability, namely: Classical Approach , Relative Frequency Approach , Subjective Approach.

P(A) = $\mathbf{ \frac{The Number of wanted outcomes }{The total number of Possible Outcomes}}$

### Basic Definition and Formula

• Random Event: If the repetition of an experiment occurs several times under similar conditions if it does not produce the same outcome every time but the outcome in a trial is one of the several possible outcomes, then such an experiment is called a Random event or a Probabilistic event.
• Elementary Event – The Elementary event refers to the outcome of each random event performed. Whenever the random event is performed, each associated outcome is known as an elementary event.
• Sample Space – Sample Space refers to the set of all possible outcomes of a random event. For example, when a coin is tossed, the possible outcomes are head and tail.
• Event – An event refers to the subset of the sample space associated with a random event.
• Occurrence of an Event – An event associated with a random event is said to occur if any one of the elementary events belonging to it is an outcome.

### Basic Probability Formulas

• Probability Range – 0 ≤ P(A) ≤ 1
• Rule of Complementary Events – P(AC) + P(A) =1
• Rule of Addition – P(A∪B) = P(A) + P(B) – P(A∩B)
• Disjoint Events – Events A and B are disjoint if P(A∩B) = 0
• Conditional Probability – P(A | B) = $\frac{P(A∩B)}{P(B)}$
• Bayes Formula – P(A | B) = $\frac{P(B|A). P(A)}{P(B)}$
• Independent Events – Events A and B are independent if. P(A∩B) = P(A) ⋅ P(B).

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### Question 1:

In a bag, there are 7 red marbles, 5 blue marbles, and 4 green marbles. Two marbles are randomly selected from the bag without replacement. What is the probability that both of them are blue?

Solution:

For the first draw, Probability of selecting a blue marble is 5/16 (5 blue marbles out of 16 total marbles).

After the first marble is drawn, there are 4 blue marbles left out of 15 total marbles.

For the second draw, the probability of selecting a blue marble is 4/15.

To calculate the probability of both marbles being blue => Probability = $\frac{5}{16}\times \frac{4}{15}$

Probability = $\frac{20}{24} = \frac{1}{12}$

### Question 2:

In a drawer, there are 4 black pens, 3 blue pens, and 5 red pens. A pen is drawn at random from the drawer. What is the probability that it is either black or blue?

Solution:

We have calculate the probability of drawing a black or blue pen from the drawer

The total no. of pens in the drawer is 4 black + 3 blue + 5 red = 12 pens

Probability of drawing a black pen is 4/12

Probability of drawing a blue pen is 3/12 = 1/4

Hence, The probability of drawing either a black or blue pen, we add the individual probabilities:

Probability = $\frac{4}{12} + \frac{3}{12}$
Probability = $\frac{7}{12}$

### Question 3:

In a bag, there are 3 green bulbs, 4 orange bulbs, and 5 white bulbs. A bulb is randomly picked from the bag. What is the probability of selecting either a green bulb or a white bulb?

Solution:

Total number of bulbs in the bag is 3 green + 4 orange + 5 white = 12 bulbs

Number of green bulbs is 3, and the number of white bulbs is 5

Probability of selecting either a green or a white bulb, we add the number of green bulbs and the number of white bulbs, and then divide it by the total number of bulbs.

Probability = (Number of green bulbs + Number of white bulbs) / Total number of bulbs
Probability = $\frac{3 + 5}{12}$
Probability = $\frac{8}{12}$
Probability = $\frac{2}{3}$

### Question 4:

Sylvester Stallone brought a box of balloons for a group of students. The box contains 3 balloons of Shape A, 4 balloons of Shape B, and 5 balloons of Shape C. If three balloons are randomly drawn from the box, what is the probability that all three balloons are of different shapes?

Solution:

Total No. of Balloons = 3 Balloons of Shape A +4 Balloons of Shape B + 5 Balloons of Shape C = 12

n(s)= $^{12}C_{3}=220$

n(e)= $^{3}C_{1} * ^{4}C_{1}* ^{5}C_{1} =60$

P= $\frac{60}{220}=\frac{3}{11}$

### Question 5:

Joey Tribbiani organized a rack race with two participants. The probability of the first participant winning is 2/7, and the probability of the second participant winning is 3/5. What is the probability that one of them will win?

Let’s denote:
P(A) = Probability of the first participant winning = 2/7
P(B) = Probability of the second participant winning = 3/5

The probability of both participants winning simultaneously (a tie) is zero since there can only be one winner. Therefore, the probability that one of them will win is:

P(one of them wins) = P(A) + P(B) – P(A and B)
P(one of them wins) = P(A) + P(B) – 0 (since P(A and B) = 0)
P(one of them wins) = P(A) + P(B)

Substituting the given probabilities:
P(one of them wins) = $\frac{2}{7}+\frac{3}{5}$
P(one of them wins) =$\frac{10}{35}+\frac{21}{35}$
P(one of them wins) =$\frac{31}{35}$

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### 2 comments on “Probability Formulas”

• Be124

*Conditional Probability – P(A | B) = {P(A∩B)}/{P(B)}