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Probability questions are a type of questions that can provide you valuable marks in competitive exams. These questions carry a weightage of 1-2 questions(2-4 marks) in SSC exams and 1-2 questions(1-2 marks) in bank exams. To perform well in competitive exams, you should clear your concepts of probability and have a good practice as there are a variety of questions.

Here are some tips for solving Probability questions: Understand basics of various concepts like dice, coin, cards etc. You should have a good hold on conepts of permutation and combinations, read carefully the question and evaluate all the possibilites. 

So, we have attached 10 questions of Probability for you to practice with. You should aim to solve these questions in less than half a minute for each.

 

<h2>Practice Questions on Probability</h2>

You can also download the Probability questions and answers pdf. Just click on the Download PDF button. So let&#39;s start with the very first question.

Q:1 Find the probability of getting a sum of at least 8 while throwing 2 dice.

1. 11/36

2. 7/36

3. 1/3

4. 2/9

(Difficulty: 2, Estimated Time: 15 Seconds) This was an easy one, isn&#39;t it?

Q:2 What is the probability of getting a co-prime when two dices are thrown?

1. 13/36

2. 17/36

3. 31/36

4. 23/36

(Difficulty: 3, Estimated Time: 20 Seconds) Sharpen your concepts to solve such questions...

Q:3 In a general survey of 832 people, it was found that 624 owned a car. If a person is selected randomly, what is the probability that the person will not be an owner of a car?

1. 1.33

2. 0.25

3. 0.75

4. 0.40

(Difficulty: 2, Estimated Time: 15 Seconds) This was a calk walk....

Q:4 A bag contains 3 red, 5 white and 4 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?

1. 23/29

2. 31/39

3. 17/27

4. 14/33

(Difficulty: 3, Estimated Time: 20 Seconds) This is a common type asked in exams..

Q:5 A bag contains 3 white balls and 5 red balls. If two balls are taken out without replacement and the first ball was found to be white, then what is the probability that second ball is red?

1. 5/8

2. 15/56

3. 5/7

4. 3/8

(Difficulty: 3, Estimated Time: 20 Seconds) We&#39;re halfway through. Have you got all your questions correct so far?

Q:6 A survey of 250 students of a school was conducted and it was found that 145 students like coffee and 105 students dislike it. Out of these students, one student is selected at random, what is the probability that the selected student like coffee?

1. 0.58

2. 0.42

3. 1

4. None of these

(Difficulty: 2, Estimated Time: 15 Seconds) You should not spend more than 10 seconds in attempting this question...

Q:7 The probability that A is telling the truth is 2/5 and of B telling the truth is 4/5. Find the probability that both will contradict each other in a situation.

1. 11/25

2. 12/25

3. 13/25

4. 14/25

(Difficulty: 4, Estimated Time: 25 Seconds) This was a bit hard. Did you get it right?

Q:8 A box contains 39 cards numbered from 1 to 39. Only 1 number is numbered on each card. While choosing a card, what is the probability of selecting a card whose face number is divisible by 3 but not by 6 or 12?

1. 5/39

2. 7/39

3. 11/39

4. 17/39

(Difficulty: 3, Estimated Time: 20 Seconds) This was not an easy one...be prepared for such questions

Q:9 In a bag, there are 3 green balls, 5 white balls, and 8 orange balls. What is the minimum number of balls that must be drawn such that there are at least four balls of the same color?

1. 8

2. 9

3. 10

4. 11

(Difficulty: 3, Estimated Time: 20 Seconds) This is a similar one like you have solved....

Q:10 There are 4 red pens, and 7 green pens. If one randomly draws a pen what is the probability of getting a red pen?

1. 3/2

2. 4/11

3. 5/9

4. 2/7

(Difficulty: 2, Estimated Time: 15 Seconds) Did you guess them all correctly?

<h2>Answer Key</h2>

Let&#39;s check out your score in this test.

<table border="0" cellpadding="1" cellspacing="1">
	<tbody>
		<tr>
			<td>1. (2)</td>
			<td>2. (4)</td>
			<td>3. (2)</td>
			<td>4. (4)</td>
			<td>5. (3)</td>
		</tr>
		<tr>
			<td>6. (1)</td>
			<td>7. (4)</td>
			<td>8. (2)</td>
			<td>9. (3)</td>
			<td>10. (2)</td>
		</tr>
	</tbody>
</table>

Comment below your score, considering each question has 1 mark only. If you scored 8 to 10, congratulations! You are one step closer to selection. If you have scored 5 to 8 marks, then you are doing well, keep it up. If you have scored less than 5 marks then you need to work a little harder on this subject. But don&#39;t worry, we are here to help you master the subject.

Let&#39;s check the answers and solutions and try to find out what went wrong. 
 

<h2>Answers and Solutions</h2>

Q:1 The correct answer is option 2 i.e. 7/36.

Sample set on throwing two dices = (1, 1) (1, 2)...... (6, 6) = 36

Favourable cases (sum of 8 or more) = (4, 4) (4, 5) (5, 4) (5, 5) (5, 6) (6, 5) (6, 6) = 7

Probability = Number of favourable cases/sample set

Probability = 7/36

 

Q:2 The correct answer is option 4 i.e. 23/26.

Two numbers are coprime if their highest common factor (or greatest common divisor ) is 1

Here we have two dice, so the total number of outcomes = 6 × 6 = 36

Here are the favorable outcomes of getting coprime

(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,3), (2,5), (3,1), (3,2), (3,4), (3,5), (4,1), (4,3), (4,5), (5,1), (5,2), (5,3), (5,4), (5,6), (6,1), (6,5)

Number of favorable outcomes = 23

∴ Probability of getting a coprime = 23/36

 

Q:3 The correct answer is option 2 i.e. 0.25.

Total number of people  = 832

Number of people who own a car = 624

Number of people who don&#39;t have a car = 832 - 624 = 208

The probability that the person will not have a car = 208/832 = 0.25

 

Q:4 The correct answer is Option 4 i.e. 14/33.

Total number of balls = (3 + 5 + 4) = 12

Probability = n(E)/n(S)

Let S be the sample spaces

E = number of ways of drawing 2 balls other than blue

n(S) = Number of ways of drawing 2 balls out of 12

n(S) = 12C2 = (12 × 11)/2 = 66

n(E) = Number of ways of drawing two balls out of 8 balls

n(E) = 8C2 = (8 × 7)/2 = 28

P(E) = n(E)/n(S) = 28/66 = 14/33

 

Q:5 The correct answer is Option 3 i.e. 5/7

If first ball is white, the remaining balls = 2 white and 5 red balls

P (second ball is red) = Number of red balls/Total number of balls

⇒ 5/(5 + 2)

⇒ 5/7

 

Q:6  The correct answer is Option 1 i.e. 0.58

Total number of students = 250

Number of students who like coffee = 145

Number of students who dislike coffee = 105

So, P (random that selected like coffee) = 145/250 = 0.58

 

Q:7 The correct answer is option 4 i.e. 14/25

Probability of A telling truth = P(A) = 2/5

Probability of B telling truth = P(B) = 4/5

Probability of A telling a lie = P(A&#39;) = 1 - (2/5) = 3/5

Probability of B telling a lie = P(B&#39;) = 1 - (4/5) = 1/5

Required Probability = P(A) × P(B&#39;) + P(B) × P(A&#39;) = (2/5) × (1/5) + (4/5) × (3/5) = (2/25) + (12/25) = 14/25

 

Q:8 The correct answer is Option 2 i.e. 7/39.

Card numbers divisible by 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39.

Card numbers that are divisible by 6 or 12 = 6, 12, 18, 24, 30, 36.

We will now remove the cards with numbers divisible by 6 or 12.

Card numbers left = 3, 9, 15, 21, 27, 33, 39

Probability = Number of favourable outcomes/Total number of outcomes

Number of favourable outcomes = 7

Total number of outcomes = 39

Probability of choosing a card on which the number marked is divisible by 3 but not by 6 or 12 = 7/39

 

Q:9 The correct answer is option 3 i.e. 10

Think the worst-case scenario :

We would pick green balls first, then pick 3 white and 3 orange balls. Now whatever ball we will pick that will be either white or orange that makes four balls of the same color, so our condition satisfies.

⇒ 3 + 3 + 3 + 1 = 10 balls

 

Q:10 The correct answer is Option 2 i.e. 4/11.

Probability = (No. of Successful Attempts / No. of total Attempts)

4 red pen & 7 green pen hence total of 11 pens

If we draw 1 pen out of the bag randomly the no. of attempts = 11C1 = No. of total Attempts

No of Successful Attempts = Attempts where a red pen is drawn out = 4C1

Probability = (4C1/11C1) = 4/11

So, this is it for today. We will meet again with another new topic. Till then, you can practice the questions again by downloading the PDF of Probability.

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