1 Marks Question - Maths STD 10 Questions - Vidyadip

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33 questions · self-marked practice — reveal the answer and mark yourself.

Question 11 Mark

In a simultaneous throw of pair of dice, find the probability of getting:
A doublet of odd numbers.

Answer

We know that a die has 6 numbers: 1, 2, 3, 4, 5, 6 and
$\therefore$ n = 6 × 6 = 36
Favourable events i.e. getting the same odd number on both the dice are
(1, 1), (3, 3) and (5, 5)
Hence total number of favourable events i.e. the same odd number on both the dice is 3
We know that $\text{Probability}=\frac{\text{Number of favourable event}}{\text{Total number of event}}$
Hence probability of getting the same odd number on both the dice is $\frac{3}{36}=\frac{1}{12}$

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Question 21 Mark

In a simultaneous throw of pair of dice, find the probability of getting:A sum greater than 9.

Answer

We know that a die has 6 numbers: 1, 2, 3, 4, 5, 6 and
$\therefore$ n = 6 × 6 = 36
A sum greater than 9, these can be (4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)
$\therefore$ m = 6
$\therefore\ \text{P(A)}=\frac{\text{m}}{\text{n}}=\frac{6}{36}=\frac{1}{6}$

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Question 31 Mark

In a simultaneous throw of pair of dice, find the probability of getting:
At least once.

Answer

We know that a die has 6 numbers: 1, 2, 3, 4, 5, 6 and
$\therefore$ n = 6 × 6 = 36
Favourable event at least once
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (3, 1), (4, 1), (5, 1), (6, 1),
Hence total number of favourable events i.e. at least once is 11
We know that $\text{Probability}=\frac{\text{Number of favourable event}}{\text{Total number of event}}$
Hence probability of getting at least once is equal to $=\frac{11}{36}$

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Question 41 Mark

In a simultaneous throw of pair of dice, find the probability of getting:
A doublet.

Answer

We know that a die has 6 numbers: 1, 2, 3, 4, 5, 6 and
$\therefore$ n = 6 × 6 = 36
Favourable events i.e. getting the same number on both the dice
(1, 1), (2, 2), (3, 3) (4, 4), (5, 5), (6, 6)
Hence total number of favourable events i.e. the same number on both the dice is 6
We know that $\text{Probability}=\frac{\text{Number of favourable event}}{\text{Total number of event}}$
Hence probability of getting the same number on both the dice $=\frac{6}{36}=\frac{1}{6}$

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Question 51 Mark

In a simultaneous throw of pair of dice, find the probability of getting:
A sum more than 7.

Answer

We know that a die has 6 numbers: 1, 2, 3, 4, 5, 6 and
$\therefore$ n = 6 × 6 = 36
A sum more than 7 which can be

(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)

which are 15.
$\therefore$ m = 15
$\therefore\ \text{P(A)}=\frac{\text{m}}{\text{n}}=\frac{15}{36}=\frac{5}{12}$

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Question 61 Mark

Fill in the blanks:
Every elementary event associated to a random experiment has ......... probability.

Answer

Every elementary event associated to a random experiment has equal probability.

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Question 71 Mark

In a simultaneous throw of pair of dice, find the probability of getting:
Even number on each die.

Answer

We know that a die has 6 numbers: 1, 2, 3, 4, 5, 6 and
$\therefore$ n = 6 × 6 = 36
Favourable outcomes for getting even number on each die are

(2,2), (2,4), (2,6),

(4,2), (4,4), (4,6),

(6,2), (6,4), (6,6)

Hence, the number of favourable outcomes are 9.
$\therefore$ P(even number on each die)
$=\frac{\text{Favourable number of ourcomes}}{\text{Total number of outcomes}}$
$=\frac{9}{36}=\frac{1}{4}$

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Question 81 Mark

In a simultaneous throw of pair of dice, find the probability of getting:
A sum less than 6.

Answer

We know that a die has 6 numbers: 1, 2, 3, 4, 5, 6 and
$\therefore$ n = 6 × 6 = 36
A sum less than 6 : These can be

(1, 1), (1, 2), (1, 4),

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

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

(4, 1)

there are 10
$\therefore$ m = 10
$\therefore\ \text{P(A)}=\frac{\text{m}}{\text{n}}=\frac{10}{36}=\frac{5}{18}$

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Question 91 Mark

In a simultaneous throw of pair of dice, find the probability of getting:An even number on first.

Answer

We know that a die has 6 numbers: 1, 2, 3, 4, 5, 6 and
$\therefore$ n = 6 × 6 = 36
Favourable events i.e. getting an even number on the first dice

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

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

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

Hence total number of favourable events i.e. getting an even number on the first dice = 18
We know that $\text{Probability}=\frac{\text{Number of favourable event}}{\text{Total number of event}}$
Hence probability of getting the an even number on the first dice $=\frac{18}{36}=\frac{1}{2}$

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Question 101 Mark

In a simultaneous throw of pair of dice, find the probability of getting:
5 as the sum.

Answer

We know that a die has 6 numbers: 1, 2, 3, 4, 5, 6 and
$\therefore$ n = 6 × 6 = 36
Sum = 5
$\therefore$ Actual events 1 + 4, 2 + 3, 3 + 2, 4 + 1 = 4
$\therefore\ \text{Probability P(E)}=\frac{4}{36}=\frac{1}{9}$

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Question 111 Mark

Fill in the blanks:
Porbability of a sure event is .....

Answer

Probability of a sure event is 1 as it is certain that it will occur always.

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Question 121 Mark

In a simultaneous throw of pair of dice, find the probability of getting:
A number other than 5 on any dice.

Answer

We know that a die has 6 numbers: 1, 2, 3, 4, 5, 6 and
$\therefore$ n = 6 × 6 = 36
A number other than 5 on the dice numbers which have 5 an be
(1, 5), (2, 5), (3, 5), (4, 5), (6, 5),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
Which are 11
$\therefore$ Number which have other that 5 will be = 36 - 11 = 25
$\therefore\ \text{P(A)}=\frac{\text{m}}{\text{n}}=\frac{25}{36}$

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Question 131 Mark

A black die and a white die are thrown at the same time. Write all the possible outcomes. What is the probability?
That the sum of the two numbers that turn up is $7$?

Answer

Given: A pair of dice is thrown
To Find: Probability of the following:
Let us first write the all possible events that can occur
$(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),$
$(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),$
$(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),$
$(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),$
$(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),$
$(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)$
Hence total number of events is $6^2=36$
Favorable events i.e. getting the sum of numbers on the dice equal to $8$
$(2,6),(3,5),(4,4),(5,3),(6,2)$
Hence total number of favorable events i.e. the sum of numbers on the dice equal to $8$ is $5$ We know that Probability $=\frac{\text { Number of favourable event }}{\text { Total number of event }}$
Hence probability of getting the sum of numbers on the dice equal to $8=\frac{5}{36}$

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Question 141 Mark

A black die and a white die are thrown at the same time. Write all the possible outcomes. What is the probability?
Of obtaining a total more than $9$?

Answer

Given: A pair of dice is thrown
To Find: Probability of the following:
Let us first write the all possible events that can occur
$(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),$
$(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),$
$(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),$
$(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),$
$(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)$
$(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)$
Hence total number of events is $6^2=36$
Favorable events i.e. getting the sum of numbers on the dice is greater than $10$ is $(5,5),(5,6),(6,4),(6,5)$ and $(6,6)$
Hence total number of favorable events i.e. getting the total of numbers on the dice greater than $9$ is $6$ We know that Probability $=\frac{\text { Number of favourable event }}{\text { Total number of event }}$
Hence probability of getting the total of numbers on the dice greater than 9 is $\frac{6}{36}=\frac{1}{6}$

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Question 151 Mark

A black die and a white die are thrown at the same time. Write all the possible outcomes. What is the probability?
Of obtaining a total of $10$?

Answer

Given: A pair of dice is thrown
To Find: Probability of the following:
Let us first write the all possible events that can occur $(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)$,
$(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)$,
$(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)$,
$(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)$,
$(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)$,
$(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)$,
Hence total number of events is $6^2=36$
Favorable events i.e. getting the sum of numbers on the dice equal to $10$ is $(4,6),(5,5)$ and $(6,4)$
Hence total number of favorable events i.e. the sum of numbers on the dice equal to $6$ is $3$
We know that Probability $=\frac{\text { Number of favourable event }}{\text { Total number of event }}$
Hence probability of getting the sum of numbers on the dice equal to $10$ is $\frac{3}{36}=\frac{1}{12}$

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Question 161 Mark

In a simultaneous throw of pair of dice, find the probability of getting:
2 will come up at least once.

Answer

We know that a die has 6 numbers: 1, 2, 3, 4, 5, 6 and
$\therefore$ n = 6 × 6 = 36
Favourable outcomes for 2 coming up atleast once are

(1, 2), (2, 1),

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

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

Hence, the number of favourable outcomes are 11.
$\therefore$ P(2 will come up at least once)
$=\frac{\text{Favourable number of outcomes}}{\text{Total number of outcomes}}=\frac{11}{36}$

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Question 171 Mark

A black die and a white die are thrown at the same time. Write all the possible outcomes. What is the probability?
That the numbers obtained have a product less then $16$.

Answer

Given: A pair of dice is thrown
To Find: Probability of the following:
Let us first write the all possible events that can occur
$(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)$,
$(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)$,
$(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)$,
$(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)$,
$(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)$,
$(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)$,
Hence total number of events is $6^2=36$
The numbers of obtained having a product less than $16=25$ i.e.,
$(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)$,
$(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)$,
$(3,1),(3,2),(3,3),(3,4),(3,5)$,
$(4,1),(4,2),(4,3)$,
$(5,1),(5,2),(5,3)$,
$(6,1),(6,2)$
$\therefore$ Probability $P ( E )=\frac{25}{36}$

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Question 181 Mark

Fill in the blanks:
The porbability of an event (other than sure and impossible event) lies between ........

Answer

Probability of an event other than sure and impossible event lies between 0 and 1.

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Question 191 Mark

In a simultaneous throw of pair of dice, find the probability of getting:
A doublet of prime numbers.

Answer

We know that a die has 6 numbers: 1, 2, 3, 4, 5, 6 and
$\therefore$ n = 6 × 6 = 36
A doublet of prime numbers: These can be
(2, 2), (3, 3), (5, 5), ⇒ m = 3
$\therefore\ \text{P(A)}=\frac{\text{m}}{\text{n}}=\frac{3}{36}=\frac{1}{12}$

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Question 201 Mark

A black die and a white die are thrown at the same time. Write all the possible outcomes. What is the probability?
That the product of numbers appearing on the top of the dice is less than $9$.

Answer

Given: A pair of dice is thrown
To Find: Probability of the following:
Let us first write the all possible events that can occur
$(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)$
$(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)$
$(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)$
$(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)$
$(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)$
$(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)$
Hence total number of events is $6^2=36$
Product of numbers appearing on the top of the dice is less that 9 , there will be
$1 \times 1,1 \times 2,1 \times 3,1 \times 4,1 \times 5,1 \times 6$
$2 \times 1,2 \times 2,2 \times 3,2 \times 4$
$3 \times 1,3 \times 2$
$4 \times 1,4 \times 2$
$5 \times 1,6 \times 1=16$
$\therefore \text { Probability } P(E)=\frac{m}{n}=\frac{16}{36}=\frac{4}{9}$

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Question 211 Mark

In a simultaneous throw of pair of dice, find the probability of getting:An even number on one and a multiple of 3 on the other.

Answer

We know that a die has 6 numbers: 1, 2, 3, 4, 5, 6 and
$\therefore$ n = 6 × 6 = 36
An even number on one and a multiple of 3 on the other
These are
(2, 3), (2, 6),
(4, 3), (4, 6),
(6, 3), (6, 6),
(3, 2), (3, 4), (3, 6),
(5, 2), (5, 4), (5, 6)
which are 11
$\therefore$ m = 11
$\text{P(A)}=\frac{\text{m}}{\text{n}}=\frac{11}{36}=\frac{11}{36}$

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Question 221 Mark

A black die and a white die are thrown at the same time. Write all the possible outcomes. What is the probability?
Of obtaining a total of $6$?

Answer

Given: A pair of dice is thrown
To Find: Probability of the following:
Let us first write the all possible events that can occur
$(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),$
$(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),$
$(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),$
$(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),$
$(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),$
$(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)$
Hence total number of events is $6^2=36$
Of obtaining a total of $6 :$ There number can be $(1,5),(2,4),(3,3),(4,2),(5,1)$ are $5$
$\therefore m=5$
$\therefore \text { Probability } P(A)=\frac{m}{n}=\frac{5}{36}$

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Question 231 Mark

A black die and a white die are thrown at the same time. Write all the possible outcomes. What is the probability?
That the sum of the two numbers appearing on the top of the dice is $13$?

Answer

Given: A pair of dice is thrown
To Find: Probability of the following:
Let us first write the all possible events that can occur

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

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

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

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

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

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

Hence total number of events is $6^2 = 36$
There is no possibility of having the sum $13$ as the sum of two greatest numbers $6$ and $6$ is $12$
$\therefore$ $m = 0$
$\therefore\ \text{P(A)}=\frac{\text{m}}{\text{n}}=\frac{0}{36}=0$

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Question 241 Mark

A black die and a white die are thrown at the same time. Write all the possible outcomes. What is the probability?
That the sum of the numbers appearing on the top of the dice is less than or equal to $12$?

Answer

Given: A pair of dice is thrown
To Find: Probability of the following:
Let us first write the all possible events that can occur

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

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

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

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

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

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

Hence total number of events is $6^2 = 36$
Favorable events i.e. getting the sum of both numbers appearing on the top of the dice less than or equal to $12$ is a sure event
Hence probability of getting the sum of both numbers appearing on the top of the dice less than or equal to $12$ is equal to $= 1$.

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Question 251 Mark

Fill in the blanks:
Sum of the probability of each outcome in an experiment is .........

Answer

Sum of the probability of each outcome in an experiment is 1.

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Question 261 Mark

In a simultaneous throw of pair of dice, find the probability of getting:
8 as the sum.

Answer

We know that a die has 6 numbers : 1, 2, 3, 4, 5, 6 and
$\therefore$ n = 6 × 6 = 36
8 as the sum: These can be (2, 6), (3, 5), (4, 4), (5, 3), (6, 2)
$\therefore$ m = 5
$\therefore\ \text{P(A)}=\frac{\text{m}}{\text{n}}=\frac{5}{36}$

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Question 271 Mark

A black die and a white die are thrown at the same time. Write all the possible outcomes. What is the probability?
Of obtaining the same number on both dice?

Answer

Given: A pair of dice is thrown
To Find: Probability of the following:
Let us first write the all possible events that can occur

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

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

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

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

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

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

Hence total number of events is $6^2 = 36$
Of obtaining the same number on both dice
There can be $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5)$ and $(6, 6)$ which are $6$
$\therefore$ $m = 6$
$\therefore\ \text{Probability P(A)}=\frac{\text{m}}{\text{n}}=\frac{6}{36}=\frac{1}{6}$

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Question 281 Mark

A black die and a white die are thrown at the same time. Write all the possible outcomes. What is the probability?
That the difference of the numbers appearing on the top of two dice is $2$.

Answer

Given: A pair of dice is thrown
To Find: Probability of the following:
Let us first write the all possible events that can occur

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

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

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

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

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

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

Hence total number of events is $6^2 = 36$
Favourable outcomes for getting the difference of the numbers as $2$ are
$(1, 3), (3, 1), (2, 4), (4, 2), (3, 5), (5, 3), (4, 6), (6, 4)$
Thus, the number of favourable outcomes are $8$.
$\therefore$ P(getting the difference of the numbers as $2$)
$=\frac{\text{Favourble number of outcomes}}{\text{Total number of outcomes}}$
$=\frac{8}{36}=\frac{2}{9}$

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Question 291 Mark

In a simultaneous throw of pair of dice, find the probability of getting:
2 will not come either time.

Answer

We know that a die has 6 numbers: 1, 2, 3, 4, 5, 6 and
$\therefore$ n = 6 × 6 = 36
2 will not come either time
$\therefore$ Number of total events = 6 × 6 = 36
= 36 - 11 = 25
$\therefore\ \text{Probability}=\frac{25}{36}$

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Question 301 Mark

In a simultaneous throw of pair of dice, find the probability of getting:
A sum less than 7.

Answer

We know that a die has 6 numbers: 1, 2, 3, 4, 5, 6 and
$\therefore$ n = 6 × 6 = 36
Favourable event a sum less than 7

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

(2, 1), (2, 2), (2, 3), (2, 4),

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

(4, 1), (4, 2), (5, 1),

Hence total number of favourable events i.e. sum less than 7 is 15
We know that $\text{Probability}=\frac{\text{Number of favourable event}}{\text{Total number of event}}$
Hence probability of getting sum less than 7 is equal to $\frac{15}{36}=\frac{5}{12}$

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Question 311 Mark

In a simultaneous throw of pair of dice, find the probability of getting:
Neither 9 nor 11 as the sum of the numbers on the faces.

Answer

We know that a die has 6 numbers: 1, 2, 3, 4, 5, 6 and
$\therefore$ n = 6 × 6 = 36
Favourable event neither 9 nor 11 as the sum of the number of faces are

(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),

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

(5, 1), (5, 2), (5, 3), (5, 5),

(6, 1), (6, 2), (6, 4), (6, 6)

Hence total number of favourable events i.e. getting neither 9 nor 11 as the sum of the number of faces is 30
We know that $\text{Probability}=\frac{\text{Number of favourable event}}{\text{Total number of event}}$
Hence probability of getting neither 9 nor 11 as the sum of the number of faces is equal to $\frac{30}{36}=\frac{5}{6}$

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Question 321 Mark

Fill in the blanks:
Porbability of an impossible event is .....

Answer

Probability of impossible event is 0 as it will never occur.

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Question 331 Mark

Fill in the blanks:
Probability of an event A + Probability of event 'not A' = ........

Answer

Probability of an event A + Probability of event 'not A' = 1.

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