3 Types to Solve Probability Problems in Easy Way

3 Types to Solve Probability Problems in Easy Way

In every competitive exam, Probability part is there. Probability is one of the easiest chapters in aptitude part. Now a day any entrance exam like bank exam, govt exam and any campus recruitment etc exam, they will ask some questions from Probability. This chapter is very interesting and somewhat difficult to prepare. Here we are providing three types of easy solving methods on Probability with examples. Candidates who are applying and appear for the any exam related to aptitude test, those can download the following 3 types to solve Probability Problems in easy way at our website. Interested aspirants may follow the below Probability examples and formulas. These probability examples may useful to your preparation.


Details about 3 Types to Solve Probability Problems in Easy Way:

Type I: Simple Problems Based on Dice, Coins, etc.

You will be very familiar with this type of problems even from school days. This type of problems is very easy to solve. Below is an example.


In a simultaneous throw of two dice, find the probability of getting a total more than 6.

Let S denote the set of all possible outcomes. S is also called sample space.

Note: If you are not clear on what ‘outcome’ means, here is an example. Let us assume that first dice shows 1 and second dice shows 1. Then the outcome is (1, 1). If the first dice shows 1 and second dice 2, then the outcome is (1, 2) and so on…

When two dice are thrown, you will get the below possible outcomes.
S = {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),
(4,1),( 4,2),( 4,3),( 4,4),( 4,5),( 4,6),
(5,1),( 5,2),( 5,3),( 5,4),( 5,5),( 5,6),
(6,1),( 6,2),( 6,3),( 6,4),( 6,5),( 6,6)}
Therefore, total number of outcomes, n(S) = 36.

Let E be the event of getting a total more than 6. In other words, sum of the numbers shown on dices should be greater than 6.
If you closely observe the values in S, you can identify outcomes where the total is greater than 6. Such outcomes will form E.

Therefore, E = {(1,6), (2,5), (2,6), (3,4), (3,5), (3,6), (4,3), (4,4), (4,5), (4,6), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
Total number of events, n(E) = 21

Probability of an even E is given by the formula, P (E) = n (E) / n(S)

Probability of getting total more than 6, P (E) = n (E) / n (S) = 21/36 = 7/12

Type II: Probability Problems that Require Logical Thinking

This is slightly difficult type you will see in bank exams. You will understand this type after reading the below example.


In a competitive exam, Arun can answer 75% questions correctly and Vimal can answer 70% questions correctly. Assume that a teacher picks a question and asks both Arun and Vimal to solve that question. what is the probability that at least one of them will solve that question?

Here, total percentage of questions, n(S) = 100
Let A be the event that Arun answers correctly and B be the event that Vimal answers correctly.

Percentage of questions Arun answers correctly, n (A) = 75
Percentage of questions Vimal answers correctly, n (B) = 70

Step 1:

Calculate Individual Probabilities
Probability of Arun answering a question correctly, P (A) = n (A)/n(S) = 75/100 = ¾
Probability of Vimal answering a question correctly, P (B) = n (B)/n(S) = 70/100 = 7/10

Probability of Arun answering a question incorrectly, P(Aꞌ) = 1 – P(A) = 1 – ¾ = ¼
Probability of Vimal answering a question incorrectly, P(Bꞌ) = 1 – P(B) = 1 – 7/10 = 3/10

Step 2: Think Logically, Read the Question Carefully and Construct the Event

Now it’s time for you to think logically. Teacher gives a question to both. For the condition that “at least one of them have to answer the question correctly”, you can construct the even as shown below.

Event E:
Arun answers correctly AND Vimal answers incorrectly OR
Arun answers incorrectly AND Vimal answers correctly OR
Both Arun and Vimal answer correctly

Based on the above event, you can write the below probability equation.
P (E) = P(A) AND P(Bꞌ)
OR P(A) And P(B)

In such probability equations, you can replace “AND” with “X” (multiplication) and “OR” with “+” (addition)
Therefore, P(E) = P(A) X P(Bꞌ) + P(Aꞌ) X P(B) + P(A) X P(B)
= (¾ x 3/10) + (1/4 x 7/10) + (¾ x 1/4)
= 9/40 + 7/40 + 3/16
= (18+14+15)/80
= 47/80

Type III: Probability Problems Based on Drawing Balls at Random

In this type, you will find a collection of balls of 2 or 3 different colors. You have to calculate the probability of drawing (picking) balls based on conditions (which you will find in question). Below example will help you to understand well.


A box contains 5 yellow and 3 green balls. Two balls are drawn at random. Find the probability that they are of the same color.

Step 1: Find n(S)
Let S be the sample space. Then, n(S) = number of ways of selecting any 2 balls.
Number of ways of selecting 2 balls from 8 balls = 8C2

(Above you can see that we have used combinations formula to calculate n(S). If you are not clear about permutations and combinations

Therefore, n(s) = 8C2 = 8 x 7 / 1 x 2 = 28
If you are not clear why 8C2 is simplified as 8X7/1X2,

Step 2: Find n (E)
Let E be the event of getting two balls of the same color.
Therefore, n(E) = number of ways of getting (2 balls out of 5 yellow) OR (2 balls out of 3 green)
Or, n(E) = 5C2 + 3C2
= (5×4 / 1×2) + (3×2 / 1×2)
= 10 + 3 = 13

Step 3: Find P (E)
Therefore, required answer = P(E) = n(E) / n(S)
= 13/28

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