Dear Reader, below are 3 important types of probability problems for bank exams. 1^{St} and 3^{rd} types are very easy, but you have to apply logical thinking for the 2^{nd} type.

At the end of the tutorial, don’t forget to take the online test.

## 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.

**Example Question 1:** In a simultaneous throw of two dice, find the probability of getting a total more than 6.

**Solution:**

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),

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),

(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.

**Example Question 2:** 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?

**Solution:**

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

**Important Note:**

You have found the probabilities of Arun and Vimal answering **correctly**. To solve this problem, you also have to find the probabilities to answer **incorrectly**. (You will be using these values in later steps.)

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 its 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(Aꞌ) AND P(B)

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 colours. 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.

**Example Question 3:** 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.

**Solution:**

**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, please read this tutorial before proceeding further.)

Therefore, n(s) = 8C2 = 8 x 7 / 1 x 2 = 28

If you are not clear why 8C2 is simplified as 8X7/1X2, please view this video before reading further

**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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