Probability MCQ Quiz - Objective Question with Answer for Probability - Download Free PDF

Last updated on Jul 2, 2025

Probability is that branch of mathematics that helps you to predict, how likely is an event to occur. Probability covers a major chunk of weightage in various entrance exams such as Banking, State CET, Aptitude Tests, etc. Testbook brings in a variety of probability questions with its solutions and explanations. It also assists you with some tips tricks to solve probability MCQs Quiz with ease. Read this article to ace your probability preparation.

Latest Probability MCQ Objective Questions

Probability Question 1:

A bag contains 3 Blue balls and 4 Black balls. What is difference between the probability of selecting 2 blue balls and the probability of selecting 3 black balls from the bag?

  1. 1/35
  2. 2/35
  3. 1/33
  4. 4/33
  5. 7/33

Answer (Detailed Solution Below)

Option 1 : 1/35

Probability Question 1 Detailed Solution

Given:

Blue balls = 3

Black balls = 4

Total balls = 3 + 4 = 7

Formula used:

Probability = Favourable outcomes ÷ Total outcomes

nCr = n! / [r!(n−r)!]

Calculations:

Ways to choose 2 blue balls = 3C2 = 3

Ways to choose any 2 balls from 7 = 7C2 = 21

⇒ P(2 blue) = 3 ÷ 21 = 1/7

Ways to choose 3 black balls = 4C3 = 4

Ways to choose any 3 balls from 7 = 7C3 = 35

⇒ P(3 black) = 4 ÷ 35

Difference = 1/7 - 4/35 = (5 - 4)/35 = 1/35

∴ The difference in probabilities is 1/35.

Probability Question 2:

Find the probability of getting a number greater than 3 in rolling of a dice once:

  1. 1/2
  2. 1/3
  3. 1/4
  4. 1/5

Answer (Detailed Solution Below)

Option 1 : 1/2

Probability Question 2 Detailed Solution

Given:

A die has 6 faces with numbers 1, 2, 3, 4, 5, and 6.

We need to find the probability of getting a number greater than 3 when rolling a die once.

Formula Used:

Probability = (Number of favorable outcomes) / (Total number of outcomes)

Calculation:

Total number of outcomes = 6 (Numbers on the die: 1, 2, 3, 4, 5, 6)

Favorable outcomes (numbers greater than 3) = {4, 5, 6}

Number of favorable outcomes = 3

Probability = Number of favorable outcomes / Total number of outcomes

⇒ Probability = 3 / 6

⇒ Probability = 1 / 2

The probability of getting a number greater than 3 is 1/2.

Probability Question 3:

A and B are two independent events such that P(A ∪ B' ) = 0.8, and P(A) = 0.3 , then P(B) =

  1. 2/7
  2. 2/3
  3. 3/8
  4. 1/8

Answer (Detailed Solution Below)

Option 1 : 2/7

Probability Question 3 Detailed Solution

Given:

P(A ∪ B') = 0.8

P(A) = 0.3

Formula used:

P(A ∪ B') = P(A) + P(B') - P(A ∩ B')

P(B') = 1 - P(B)

Calculations:

P(A ∪ B') = P(A) + (1 - P(B)) - P(A ∩ B')

Since A and B are independent, P(A ∩ B') = P(A) × P(B')

⇒ 0.8 = P(A) + 1 - P(B) - P(A) × P(B')

⇒ 0.8 = P(A) + 1 - P(B) - P(A) × (1 - P(B))

⇒ 0.8 = 0.3 + 1 - P(B) - 0.3 × (1 - P(B))

⇒ 0.8 = 0.3 + 1 - P(B) - 0.3 + 0.3P(B)

⇒ 0.8 - 1 = -P(B) + 0.3 × P(B)

⇒ -0.2 = -P(B)(1 - 0.3)

⇒ -0.2 = -P(B) × 0.7

⇒ P(B) = 0.2 ÷ 0.7

⇒ P(B) = 2/7

∴ P(B) = 2/7

Probability Question 4:

If 4 coins are tossed once then what is the probability of getting exactly 2 heads?

  1. 7/8
  2. 5/8
  3. 1/2
  4. 3/8

Answer (Detailed Solution Below)

Option 4 : 3/8

Probability Question 4 Detailed Solution

Given:

Number of coins tossed = 4

Number of heads required = 2

Each coin has two outcomes: Head (H) or Tail (T).

Formula Used:

Probability of an event = (Number of favorable outcomes) / (Total number of outcomes)

Total outcomes for tossing n coins = 2n

Number of ways to get r heads in n tosses = nCr = n! / [r!(n - r)!]

Calculation:

Total number of outcomes = 24 = 16

Number of ways to get exactly 2 heads = 4C2 = 4! / [2!(4 - 2)!]

⇒ 4C2 = (4 × 3) / (2 × 1)

⇒ 4C2 = 6

Probability of getting exactly 2 heads = (Number of favorable outcomes) / (Total number of outcomes)

⇒ Probability = 6 / 16

⇒ Probability = 3 / 8

The probability of getting exactly 2 heads is 3/8.

Probability Question 5:

A single letter is drawn at random from the word "ASPIRATION" the probability that it is a vowel is:

  1. 1/2
  2. 1/3
  3. 1/4
  4. 0

Answer (Detailed Solution Below)

Option 1 : 1/2

Probability Question 5 Detailed Solution

Given:

Word = ASPIRATION

Total letters = 10

Vowels = A, I, A, I, O

Total vowels = 5

Formula Used:

Probability = Number of favorable outcomes / Total outcomes

Calculation:

Number of favorable outcomes = 5

Total outcomes = 10

Probability = 5 / 10

Probability = 1 / 2

Solution:

The probability that the letter drawn is a vowel is 1/2.

Top Probability MCQ Objective Questions

Probability of 3 students solving a question are  and . Probability to solve the question is:

Answer (Detailed Solution Below)

Option 4 :

Probability Question 6 Detailed Solution

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Probability of 3 students,

P(A) = 1/2, P(A̅) = 1/2

P(B) = 1/3, P(B̅) = 2/3

P(C) = 1/4, P(C̅) = 3/4

So, Probability of no one solve the question is =  = 1/4

⇒ P(None) = 1/4

Then, The probability to solve the question is = 1 - 1/4 = 3/4

Hence, the correct answer is "3/4".

Two cards are drawn randomly from a pack of 52 cards. What is the probability of getting one spade card and one diamond card?

  1. 13/51
  2. 1/2
  3. 1/4
  4. 13/102

Answer (Detailed Solution Below)

Option 4 : 13/102

Probability Question 7 Detailed Solution

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Concept used:

Total number of spade cards = 13

Total number of diamond cards = 13

Formula used:

P = Favorable outcomes/Total outcomes 

Calculation:

Total outcomes = 52C2 =  =  = 1326

Favourable outcomes = 13C1 × 13C1

= 13 × 13 = 169

∴ The Required Probability = 169/1326 = 13/102

Probability of getting a multiple of 2 on one dice and a multiple of 3 on the other dice, when both dice are thrown simultaneously is:

Answer (Detailed Solution Below)

Option 3 :

Probability Question 8 Detailed Solution

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Given:

Favorable outcomes are a multiple of 2 on one dice and a multiple of 3 on the other dice.

Concept used:

When two dice are thrown total outcome = 6 × 6 = 36

Probability = favourable outcomes/total outcomes

Explanation:

Favourable outcomes = (2,3), (4,3), (6,3), (2,6), (4,6), (6,6),

(3,2), (3,4), (3,6), (6,2), (6,4) = 11

Total outcomes = 6 × 6 = 36

∴ Probability = 11/36

The sample space of four coins tossed together is:

  1. 8
  2. 64
  3. 32
  4. 16

Answer (Detailed Solution Below)

Option 4 : 16

Probability Question 9 Detailed Solution

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Number of coins tossed = 4

∴ Sample space of four coins tossed = 24 = 16

Ajay rolled two dice together. What is the probability that first dice showed a multiple of 3 and the second dice showed an even number?

Answer (Detailed Solution Below)

Option 1 :

Probability Question 10 Detailed Solution

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GIVEN:

One dice shows a multiple of 3.

Other dice shows even number.

CONCEPT:

Total number of outcomes in two dice is 36.

FORMULA USED:

P = Favorable outcomes/Total outcomes 

CALCULATION:
There are only 6 such cases as required,

(3,2), (3,4) (3,6) (6,2) (6,4) (6,6)

∴ Required probability = 6/36 = 1/6

∴ The probability is 1/6.

The probability of Sita, Gita and Mita passing a test is 60%, 40% and 20% respectively. What is the probability that at Sita and Gita will pass the test and Mita will not?

  1. 38.4%
  2. 60%
  3. 4.8%
  4. 19.2%

Answer (Detailed Solution Below)

Option 4 : 19.2%

Probability Question 11 Detailed Solution

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Given

Probability of passing the test by Sita = 60% = 60/100

Probability of passing the test by Gita = 40% = 40/100

Probability of passing the test by Mita = 20% = 20/100

Formula

Probability of not happening even A = 1 - Probability of  happening even A

Probability of happening A and B = Probability of happening A × Probability of happening B

Calculation

Probability of not passing the test by Mita = 1 - Probability of passing the test by Mita

= 1 - (20/100)

= 80/100

Now,

Probability that at Sita and Gita will pass the test and Mita will not = Probability of passing the test by Sita × Probability of passing the test by Gita × Probability of not passing the test by Mita

= (60/100) × (40/100) × (80/100)

= 192/1000

= (192/10)%

= 19.2%

A bag contains 5 black and 6 white balls; two balls are drawn at random. What is the probability that the balls drawn are white?

  1. 10/11
  2. 4/11
  3. 6/11
  4. 3/11

Answer (Detailed Solution Below)

Option 4 : 3/11

Probability Question 12 Detailed Solution

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Given

Number of black balls = 5

Number of white balls = 6

Formula

Probability = Favorable events/Total possible events

Calculation

Favorable event = 6C2

Total possible events = 11C2

∴ Probability = 6C2/11C2 = (6 × 5)/(11 × 10) = 3/11

Three coins are tossed simultaneously. Find the probability of getting exactly two heads.

  1. 5/8
  2. 1/8
  3. 1/2
  4. 3/8

Answer (Detailed Solution Below)

Option 4 : 3/8

Probability Question 13 Detailed Solution

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Given:

Three coins are tossed simultaneously.

Formula:

Probability = Number of favorable outcomes/ Total number of outcomes.

Calculation:

When three coins are tossed then the outcome will be any one of these combinations. (TTT, THT, TTH, THH. HTT, HHT, HTH, HHH).

So, the total number of outcomes is 8.

Now, for exactly two heads, the favorable outcome is (THH, HHT, HTH).

We can say that the total number of favorable outcomes is 3.

Again, from the formula

Probability = Number of favorable outcomes/Total number of outcomes

Probability = 3/8

The probability of getting exactly two heads is 3/8.

A dice is rolled two times. Find the probability of getting a composite number on first roll and a prime number on second roll?

Answer (Detailed Solution Below)

Option 2 :

Probability Question 14 Detailed Solution

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Calculation

Number of composite number in a dice are  (4 and 6)

⇒ Probabilty of composite number in a dice = 2/6 = 1/3

⇒ Number of prime number in a dice = 2, 3 and 5

⇒ Probability of prime number in a dice = 3/6 = 1/2

∴The probability of getting a composite number on the first roll and a prime number on the second roll = 1/2 × 1/3 = 1/6

Two unbiased dice are rolled simultaneously. Find the probability of getting sum greater than 5.

  1. 13/18
  2. 23/36
  3. 7/9
  4. 5/9
  5. 11/18

Answer (Detailed Solution Below)

Option 1 : 13/18

Probability Question 15 Detailed Solution

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GIVEN:

Number of unbiased dice = 2

CONCEPT:

Probability (Event) = Number of favorable outcome / Total outcome

CALCULATION:

No. of ways of rolling a pair of dice = 6 × 6 = 36

Let E = event of getting a sum greater than 5 = {(1, 6), (1, 5), (2, 6),(2, 5), (2, 4), (3, 6), (3, 5), (3, 4), (3, 3), (4, 6), (4, 5), (4, 4), (4, 3), (4, 2), (5, 6), (5, 5),(5, 4), (5, 3), (5, 2), (5, 1), (6, 6), (6, 5), (6, 4), (6, 3), (6, 2), (6,1)}

n(E) = 26

⇒ Required probability = 26/36 = 13/18

⇒ The probability of getting sum greater than 5 = 13/18

 

GIVEN:

Number of unbiased dice = 2

CONCEPT:

Probability (Event) = 1 - (Number of non favorable outcome / Total outcome)

Probability of getting a sum greater than 5 = 1 - (Probability of getting a sum less than or equal to 5)

CALCULATION:

No. of ways of rolling a pair of dice = 6 × 6 = 36

Let F = event of getting a sum less than or equal to 5 = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4,1)}

n(F) = 10

⇒ Required probability = 1 - (10/36) = 1 - (5/18) = 13/18

∴ The probability of getting sum greater than 5 = 13/18

Important Points

When we have a large number of cases like 26 in case of Event (E) then we calculate non-favorable outcome(Compliment event i.e. 1 - favourable event)

Mistake Points

In this question, we have to avoid the cases in which sum of digit is equal to five like{(1, 4),(2, 3),(3, 2),(4,1)}

Additional Information

Probabilities for the two dice

The total sum from two dice Number of combinations
2 1
3 2
4 3
5 4
6 5
7 6
8 5
9 4
10 3
11 2
12 1
Total 36

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