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

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.

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

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

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

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

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.

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.

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)  =  \(\dfrac{5}{12}\)

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 =\(\rm \frac{1}{36}\)

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.

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.

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).

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

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

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)  =  \(\dfrac{1}{12}\)