Problems on die MCQ Quiz - Objective Question with Answer for Problems on die - Download Free PDF

Last updated on Apr 11, 2025

Latest Problems on die MCQ Objective Questions

Problems on die Question 1:

Three perfect dice D1, D2 and D3 and are rolled. Let x, represent the numbers on D1, D2 and D3 respectively. What is the number of possible outcomes such that x < y < z?

  1. 20
  2. 18
  3. 14
  4. 10

Answer (Detailed Solution Below)

Option 1 : 20

Problems on die Question 1 Detailed Solution

Explanation:

Possible outcomes x < y < z

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

Thus, Total possible outcomes = 20

∴ Option (a) is correct.

Problems on die Question 2:

Three perfect dice are rolled. Under the condition that no two show the same face, what is the probability that one of the faces shown is an ace (one)?

  1. 5/9
  2. 2/3
  3. 1/3
  4. 1/2

Answer (Detailed Solution Below)

Option 4 : 1/2

Problems on die Question 2 Detailed Solution

Explanation:

⇒n(S) = 6 × 5 × 4

⇒ n(E) = n (one of the faces shown is an ace.)

=  3(1 × 5 × 4) = 3 × 5 × 4 

⇒ \(P(E) = \frac{3\times 5\times 4}{6\times 5\times4} = \frac{1}{2}\)

∴ Option (d) is correct

Problems on die Question 3:

A die has two faces with number 4, three faces with number 5 and one face with number 6. If the die is rolled once, then what is the probability of getting 4 or 5?

  1. 1/3
  2. 2/3
  3. 5/6
  4. 1/2

Answer (Detailed Solution Below)

Option 3 : 5/6

Problems on die Question 3 Detailed Solution

Explanation:

Number on faces on die are 4, 4, 5, 5, 5 and 6.

⇒n(S) = 6

P (getting 4 or 5) = 5/6

∴ Option (c) is correct

Problems on die Question 4:

One die has two faces marked 1, two faces marked 2, one face marked 3 and one face marked 4. Another die has one face marked 1, two faces marked 2, two faces marked 3 and one face marked 4. The probability of getting the sum of numbers to be 4 or 5, when both the dice are thrown together, is

  1. \(\frac{1}{2}\)
  2. \(\frac{3}{5}\)
  3. \(\frac{2}{3}\)
  4. \(\frac{4}{9}\)

Answer (Detailed Solution Below)

Option 1 : \(\frac{1}{2}\)

Problems on die Question 4 Detailed Solution

Calculation

a = number on dice 1

b = number on dice 2

(a, b) = (1,3), (3,1), (2,2), (2,3), (3,2), (1,4), (4,1)

Required probability

\(\frac{2}{6} \times \frac{2}{6}+\frac{1}{6} \times \frac{1}{6}+\frac{2}{6} \times \frac{2}{6}+\frac{2}{6} \times \frac{2}{6}+\frac{1}{6} \times \frac{2}{6}+\frac{2}{6} \times \frac{1}{6}+\frac{1}{6} \times \frac{2}{6}\)

\(\frac{18}{36}=\frac{1}{2}\)

Hence option 1 is correct

Problems on die Question 5:

A six faced die is a biased one. It is thrice more likely to show an odd numbers than show an even number. It is thrown twice. The probability that the sum of the numbers in two throws is even, is

  1. \(\frac{5}{9}\)
  2. \(\frac{5}{8}\)
  3. \(\frac{1}{2}\)
  4. None of these

Answer (Detailed Solution Below)

Option 2 : \(\frac{5}{8}\)

Problems on die Question 5 Detailed Solution

Calculation

Let P be the probability of getting an even number. Then the probability of getting an odd number is 3P.

Since the events of getting an even number and an odd number are mutually exclusive and exhaustive.

∴ P + 3P = 1 ⇒ P = 1/4

Thus, the probability of getting an odd number in a single throw is 3/4 and that of an even number is 1/4.

If the die is thrown twice, then the sum of the numbers in two throws is even if both the numbers are even or both are odd. 

∴ Required probability = \(\frac{3}{4} \times \frac{3}{4}+\frac{1}{4} \times \frac{1}{4}=\frac{10}{16}=\frac{5}{8}\)  

Hence option 2 is correct

Top Problems on die MCQ Objective Questions

If four dice are thrown together, then what is the probability that the sum of the numbers appearing on them is 25?

  1. 0
  2. 1/2
  3. 1
  4. 1/1296

Answer (Detailed Solution Below)

Option 1 : 0

Problems on die Question 6 Detailed Solution

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Concept:

Probability of an event happening = \(\rm \dfrac{\text{(Number of ways it can happen)}}{\text{ (Total number of outcomes)}}\)

If a die thrown, Number of sample space = 6, If two dice are thrown n(S) = 62 = 36

Calculation:

Here, four dice are thrown, 

n(S) = 64

Now, sum of the numbers appearing on them 25 = { }       

⇒ n = 0               

(∵maximum sum = 6 + 6 + 6 + 6 = 24)

∴ Probability = 0/(64) = 0

Hence, option (1) is correct.

What is the probability of getting a sum of 7 with two dice?

  1. 1/6
  2. 1/3
  3. 1/12
  4. 5/36

Answer (Detailed Solution Below)

Option 1 : 1/6

Problems on die Question 7 Detailed Solution

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Concept:

Probability of an event happening = (Number of ways it can happen) / (Total number of outcomes)

 

Calculation:

Two dice thrown,

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)}

 

n(S) = 36

 

Now, let A be an event getting sum 7

A = {(1, 6), (2, 5), (3, 4), (4, 3), (5,, 2), (6, 1)}

n(A) = 6,

∴Probability of getting sum 7 = n(A) /n(S)

= 6/36

= 1/6

 

Hence, option (1) is correct.

In a simultaneous throw of a pair of dice, the probability of getting a total more than 7 is

  1. \(\dfrac{7}{12}\)
  2. \(\dfrac{5}{36}\)
  3. \(\dfrac{5}{12}\)
  4. \(\dfrac{7}{36}\)

Answer (Detailed Solution Below)

Option 3 : \(\dfrac{5}{12}\)

Problems on die Question 8 Detailed Solution

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Concept:

Let S be sample space. The probability of an event is P (E) 

⇒ P (E)  = \(\rm \dfrac {n(E)}{n(S)}\)

 

Calculations:

Given, S = a simultaneous throw of a pair of dice.

{(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) }

⇒ n(S ) = 36

 E = event of getting a total more than 7 

E = {(2, 6), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6),(5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

⇒ n(E ) = 15

The probability of getting a total more than 7 is P (E) 

⇒ P (E)  = \(\rm \dfrac {n(E)}{n(S)}\)

⇒ P (E)  = \(\rm \dfrac {15}{36}\)

⇒ P (E)  =  \(\dfrac{5}{12}\)

Three dice are thrown. What is the probability that the same number will appear on each of them?

  1. 1 / 6
  2. 1 / 18
  3. 1 / 24
  4. 1 / 36

Answer (Detailed Solution Below)

Option 4 : 1 / 36

Problems on die Question 9 Detailed Solution

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Concept:

Probability is the ratio of the number of outcomes in an exhaustive set of equally likely outcomes that produce a given event to the total number of possible outcomes

Calculations:

The possible outcomes of a dice are = {1, 2, 3, 4, 5, 6} 

Three dice are thrown simultaneously.

Number of outcomes S = 63

Now, possible outcomes of three dice the same number will appear on each of them are = {(1,1, 1),(2,2,2),....(6,6,6)}

Number of outcomes of three dice with the same number will appear on each of them  A = 6

Now probability that the same number will appear on each of them = \(\rm \frac{n(A)}{n(S)}\)

probability that the same number will appear on each of them =\(\rm \frac{6}{6^3}\)

probability that the same number will appear on each of them =\(\rm \frac{1}{36}\)

In a single throw of two dice, find the probability of obtaining ‘a total of 8’

  1. \(\frac{7}{36}\)
  2. \(\frac{1}{9}\)
  3. \(\frac{1}{36}\)
  4. \(\frac{5}{36}\)

Answer (Detailed Solution Below)

Option 4 : \(\frac{5}{36}\)

Problems on die Question 10 Detailed Solution

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Concept:

In a random experiment, let S be the sample space and let E ⊆ S. Then, E is an event.

The probability of occurrence of E is defined as,

\(\rm P(E)=\frac{n(E)}{n(S)}\)where, n(E) = number of elements in E and n(S) = number of possible outcomes.

 

Calculation:

We know that in a single throw of two dice, the total number of possible outcomes is 6 × 6 = 36

Let S be the sample space. Then, n(S) = 36

Let E = event of getting a total of 8. Then, 

E = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)} 

∴  n(E) =  5

So, the probability of obtaining a total of '8’  = P(E) = \(\rm \frac{n(E)}{n(S)}=\frac{5}{36}\)

Hence, option (4) is correct.

In a single throw of two dice, find the probability of first dice always occur odd number and sum of two dice is greater than 5.

  1. \(\frac{1}{4}\)
  2. \(\frac{2}{3}\)
  3. \(\frac{3}{4}\)
  4. \(\frac{1}{3}\)

Answer (Detailed Solution Below)

Option 4 : \(\frac{1}{3}\)

Problems on die Question 11 Detailed Solution

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Concept:

In a random experiment, let S be the sample space and let E ⊆ S. Then, E is an event.

The probability of occurrence of E is defined as,

\(\rm P(E)=\frac{n(E)}{n(S)}\)where, n(E) = number of elements in E and n(S) = number of possible outcomes.

Calculation:

We know that in a single throw of two dice, the total number of possible outcomes is 6 × 6 = 36

Let S be the sample space. Then, n(S) = 36

Let E = first dice always occur odd number and sum of two dice is greater than 5. Then, 

E = {(1,5 ), (1, 6), (3, 3), (3, 4), (3, 5), (3, 6), (5,1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)} 

∴ n(E) =  12

So, the probability of obtaining a total of '12’  = P(E) = \(\rm \frac{n(E)}{n(S)}=\frac{12}{36} = {1\over3}\)

Hence, option (4) is correct.

A pair of fair dice is rolled. What is the probability that the second dice lands on a higher value than does the first?

  1. \(\frac{1}{4}\)
  2. \(\frac{1}{6}\)
  3. \(\frac{5}{{12}}\)
  4. \(\frac{5}{{18}}\)

Answer (Detailed Solution Below)

Option 3 : \(\frac{5}{{12}}\)

Problems on die Question 12 Detailed Solution

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Concept

The sample space when a pair of dice is rolled is as follows:

(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)

Total outcomes = 36

Calculation

The pairs in which the second number is greater than the first number are as follows:

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

Total favourable outcomes = 15

Required probability

\( = \frac{{15}}{{36}} = \frac{5}{{12}}\)

Correct option is (3).

A die is thrown 3 times. Events A and B stated below:

4 on the 3rd throw

6 on the 1st and 5 on the 2nd throw

What will be the probability of A knowing that B has already taken place?

  1. \(\frac{{1}}{{108}}\)
  2. \(\frac{{1}}{{36}}\)
  3. \(\frac{{1}}{{6}}\)
  4. \(\frac{{1}}{{216}}\)

Answer (Detailed Solution Below)

Option 3 : \(\frac{{1}}{{6}}\)

Problems on die Question 13 Detailed Solution

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Given:

A die is thrown 3 times.

Events A and B stated:-

4 on the 3rd throw.

6 on the 1st and 5 on the 2nd throw.

Formula Used:

P(A|B) = P(A ∩ B)/P(B)

Calculation:

A die is thrown 3 times.

Total cases = 6 × 6 × 6 =216

A : 4 on the third throw.

B : 6 on the first and 5 on the second throw.

A = {(1, 1, 4), (1, 2, 4),....(1, 6, 4), (2, 1, 4), (2, 2, 4),....(2, 6, 4),

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

        (5, 1, 4), (5, 2, 4),.....(5, 6, 4), (6, 1, 4), (6, 2, 4),....(6, 6, 4)}

⇒ P(A) = \(\frac{36}{216}\)

B = {(6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)}

⇒ P(B) = \(\frac{6}{216}\)

Thus A ∩ B = {(6, 5, 4)}

So, P(A ∩ B) = \(\frac{1}{216}\)

Now, P(A|B) = \(\frac{P(A ∩ B)}{P(B)}\)

⇒ P(A|B) = \(\frac{\frac{1}{216}}{\frac{6}{216}}\)

⇒ P(A|B) = \(\frac{1}{6}\)

∴ The probability of A knowing that B has already taken place is 1/6

If three dice are rolled under the condition that no two dice show the same face, then what is the probability that one of the faces is having the number 6?

  1. \(\frac{5}{6}\)
  2. \(\frac{5}{9}\)
  3. \(\frac{1}{2}\)
  4. \(\frac{5}{12}\)

Answer (Detailed Solution Below)

Option 3 : \(\frac{1}{2}\)

Problems on die Question 14 Detailed Solution

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Concept:

Probability of an event happening = (Number of ways it can happen) / (Total number of outcomes)

If three dice trhown, number of sample space = 63 = 216

 

Calculation:

Three dice are rolled, number of sample space = 216

Two dice show the same face =

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

(2, 2, 2).....,

(3, 3, 3)......,

(4, 4, 4)......,

(5, 5, 5)......,

(6, 6, 6).......}

n = 16 × 6 = 96

Number of no two dice show the same face, n(S) = 216 - 96 = 120

 

Faces having no number 6,

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

(2, 1, 3), .......

..........

.........

..........(5, 4, 3)}

n = 12 × 5 = 60

So the number of one of the faces is having the number 6 = 120 - 60 = 60

∴ The probability that one of the faces is having the number 6 = 60/120 = 1/2

Hence, option (3) is correct.

 

  

A die is thrown: Sample space S = {1, 2, 3, 4, 5, 6}

Now, possible outcomes of three dice the condition that no two dice show the same face = -  -  - = 6 × 5 × 4 = 120

Now, the number of one of the faces is having the number 6 = 1 × 5 × 4 × 3! = 60

∴ The probability that one of the faces is having the number 6 = 60/120 = 1/2

In a simultaneous throw of a pair of dice, the probability of getting a total more than 10 is

  1. \(\dfrac{5}{12}\)
  2. \(\dfrac{7}{12}\)
  3. \(\dfrac{1}{12}\)
  4. \(\dfrac{1}{4}\)

Answer (Detailed Solution Below)

Option 3 : \(\dfrac{1}{12}\)

Problems on die Question 15 Detailed Solution

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Concept:

Let S be sample space. The probability of an event is P (E) 

⇒ P (E)  = \(\rm \dfrac {n(E)}{n(S)}\)

 

Calculations:

Given, S = a simultaneous throw of a pair of dice.

{(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) }

⇒ n(S ) = 36

 E = event of getting a total more than 10 

E = {(5, 6), (6, 5), (6, 6)}

⇒ n(E ) = 3

The probability of getting a total more than 7 is P (E) 

⇒ P (E)  = \(\rm \dfrac {n(E)}{n(S)}\)

⇒ P (E)  = \(\rm \dfrac {3}{36}\)

⇒ P (E)  =  \(\dfrac{1}{12}\)

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