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MATHS 2 Marks Questions

Data Handling – IV (Probability) · Bihar Board (BSEB) STD 8 (BSEB - English Medium). 21 questions.

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21 questions · timed · auto-graded

Question 12 Marks
Three coins are tossed together. Find the probability of getting: no tails.
Answer
When 3 coins are tossed together, the outcomes are as follows: $S = \{(h,h,h), (h, h, t), (h, t, h), (h, t, t), (t, h, h), (t, h, t), (t, t, h), (t, t, t)\}$Therefore, the total number of outcomes is $8.$ Let $A$ be the event of getting triplets having no tail. Triplets having no tail: $(h, h, h)$Therefore, the total number of favourable outcomes is $1.$
$\text{P(A)}=\frac{\text{Number of favourable outcomesTotal}}{\text{ number of outcomes}}=\frac{1}{8}$
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Question 22 Marks
In a simultaneous throw of a pair of dice, find the probability of getting: a doublet of odd numbers.
Answer
When a pair of dice is thrown simultaneously, the sample space will be as follows: $S = \{(1, 1), (1, 2), (1, 3), (1, 4),$ ______ $(6, 5), (6, 6)\}$ Hence, the total number of outcomes is $36.$ Let A be the event of getting doublets of odd numbers in the sample space. The doublets of odd numbers in the sample space are $(1, 1), (3, 3)$ and $(5, 5).$ Hence, the number of favourable outcomes is $3.$
$∴\text{ P(A)}=\frac{\text{Number of favourable outcomesTotal}}{\text{ number of outcomes}}=\frac{3}{36}=\frac{1}{12}$
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Question 32 Marks
What is the probability that a number selected from the numbers $1, 2, 3,$ _____, $15$ is a multiple of $4?$
Answer
There are $15$ numbers from $1, 2,$ ______$, 15$. Hence, the total number of cases is $15.$ Again, the multiples of $4$ are $4, 8$ and $12.$ Therefore, the total number of favourable cases is $3.$
$\therefore ($the number is a multiple of $4)$
$=\frac{\text{Number of favourable casesTotal}}{\text{ number of cases} }=\frac{3}{15}=\frac{1}{5}$
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Question 42 Marks
An urn contains $10$ red and $8$ white balls. One ball is drawn at random. Find the probability that the ball drawn is white.
Answer
Number of red balls $= 10$
Number of white balls $= 8$
Total number of balls in the urn $= 10 + 8 = 18$
Therefore, the total number of cases is $18$ and the number of favourable cases is $8.$
$\therefore\text{P}\text{ (The ball drawn is white) }=\frac{\text{Number of favourable cases}}{\text{Total number of cases}}\\=\frac{8}{18}=\frac{4}{9}$
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Question 52 Marks
In a simultaneous throw of a pair of dice, find the probability of getting: a sum less than $7.$
Answer
When a pair of dice is thrown simultaneously, the sample space will be as follows: $S = \{(1, 1), (1, 2), (1, 3), (1, 4),$ ______ $(6, 5), (6, 6)\}$ Hence, the total number of outcomes is $36.$ Let A be the event of getting pairs whose sum is less than $7.$ The pairs whose sum is less than $7$ are $(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2)$ and $(5, 1)$.
Hence, the number of favourable outcomes is $15.$
$\therefore \text{ P(A)}=\frac{\text{Number of favourable outcomesTotal}}{\text{ number of outcomes}}$
$=\frac{15}{36}=\frac{5}{12}$
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Question 62 Marks
Three coins are tossed together. Find the probability of getting: at least two heads.
Answer
When $3$ coins are tossed together, the outcomes are as follows: $S = \{(h,h,h), (h, h, t), (h, t, h), (h, t, t), (t, h, h), (t, h, t), (t, t, h), (t, t, t)\}$Therefore, the total number of outcomes is $8.$ Let A be the event of getting triplets having at least $2$ heads. Triplets having at least 2 heads: $(h, h, t), (h, t, h), (t, h, h), (h, h, h)$Therefore, the total number of favourable outcomes is $4.$ $\text{P(A)}=\frac{\text{Number of favourable outcomesTotal}}{\text{ number of outcomes}}$$=\frac{4}{8}=\frac{1}{2}$
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Question 72 Marks
In a simultaneous throw of a pair of dice, find the probability of getting: a sum greater than $9.$
Answer
When a pair of dice is thrown simultaneously, the sample space will be as follows: $S = \{(1, 1), (1, 2), (1, 3), (1, 4),$ ______ $(6, 5), (6, 6)\}$ Hence, the total number of outcomes is $36.$ Let A be the event of getting pairs whose sum is greater than $9.$ The pairs whose sum is greater than $9$ are $(4, 6), (5, 5), (5, 6), (6, 4), (6, 5)$ and $(6, 6).$ Hence, the number of favorable outcomes is $6.$
$\therefore \text{ P(A)}=\frac{\text{Number of favourable outcomesTotal}}{\text{ number of outcomes}}=\frac{6}{36}=\frac{1}{6}$
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Question 82 Marks
In a simultaneous throw of a pair of dice, find the probability of getting: a sum less than $6.$
Answer
When a pair of dice is thrown simultaneously, the sample space will be as follows: $S = \{(1, 1), (1, 2), (1, 3), (1, 4),$ ______ $(6, 5), (6, 6)\}$ Hence, the total number of outcomes is $36.$ Let A be the event of getting pairs whose sum is less than $6.$ The pairs whose sum is less than $6$ are $(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2)$ and $(4, 1).$ Hence, the number of favourable outcomes is $10.$
$\therefore \text{ P(A)}=\frac{\text{Number of favourable outcomesTotal}}{\text{ number of outcomes}}=\frac{10}{36}=\frac{5}{18}$
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Question 92 Marks
In a simultaneous throw of a pair of dice, find the probability of getting: a sum more than $7.$
Answer
When a pair of dice is thrown simultaneously, the sample space will be as follows: $S = \{(1, 1), (1, 2), (1, 3), (1, 4),$ ______ $(6, 5), (6, 6)\}$ Hence, the total number of outcomes is $36.$ Let $A$ be the event of getting pairs whose sum is more than $7.$ The pairs whose sum is more than $7$ are $(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)$ and $(6, 6).$ Hence, the number of favourable outcomes is $15.$
$\therefore \text{ P(A)}=\frac{\text{Number of favourable outcomesTotal}}{\text{ number of outcomes}}$
$=\frac{15}{36}=\frac{5}{12}$
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Question 102 Marks
In a simultaneous throw of a pair of dice, find the probability of getting: a doublet of prime numbers.
Answer
When a pair of dice is thrown simultaneously, the sample space will be as follows: $S = \{(1, 1), (1, 2), (1, 3), (1, 4),$ ______ $(6, 5), (6, 6)\}$ Hence, the total number of outcomes is $36.$ Let A be the event of getting doublets of prime numbers in the sample space. The doublets of prime numbers in the sample space are $(2, 2), (3, 3)$ and $(5, 5).$ Hence, the number of favourable outcomes is $3.$
$\therefore \text{ P(A)}=\frac{\text{Number of favourable outcomesTotal}}{\text{ number of outcomes}}=\frac{3}{36}=\frac{1}{12}$
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Question 112 Marks
In a simultaneous throw of a pair of dice, find the probability of getting: an even number on first.
Answer
When a pair of dice is thrown simultaneously, the sample space will be as follows: $S = \{(1, 1), (1, 2), (1, 3), (1, 4),$ ______ $(6, 5), (6, 6)\}$ Hence, the total number of outcomes is $36.$ Let A be the event of getting pairs who has even numbers on first in the sample space. The pairs who has even numbers on first are: $(2, 1), (2, 2),$ _____ $(2, 6), (4, 1),$ _____ $,(4, 6), (6, 1),$ _____ $(6, 6)$. Hence, the number of favourable outcomes is $18.$
$\therefore \text{ P(A)}=\frac{\text{Number of favourable outcomesTotal}}{\text{ number of outcomes}}=\frac{18}{36}=\frac{1}{6}$
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Question 122 Marks
Three coins are tossed together. Find the probability of getting: at least one head and one tail.
Answer
When $3$ coins are tossed together, the outcomes are as follows: $S = \{(h,h,h), (h, h, t), (h, t, h), (h, t, t), (t, h, h), (t, h, t), (t, t, h), (t, t, t)\}$Therefore, the total number of outcomes is $8$. Let A be the event of getting triplets having at least one head and one tail.Triplets having at least one head and one tail: $(h, h, t), (h, t, h), (t, h, h), (h, h, t), (t, t, h), (t, h, t)$ Therefore, the total number of favourable outcomes is $6.$
$\text{P(A)}=\frac{\text{Number of favourable outcomesTotal}}{\text{ number of outcomes}}$$=\frac{6}{8}=\frac{3}{4}$
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Question 132 Marks
In a simultaneous throw of a pair of dice, find the probability of getting: a number other than $5$ on any dice.
Answer
When a pair of dice is thrown simultaneously, the sample space will be as follows: $S = \{(1, 1), (1, 2), (1, 3), (1, 4),$ ______ $(6, 5), (6, 6)\}$ Hence, the total number of outcomes is $36.$ Let A be the event of getting pairs that has the number $5$ The pairs that has the number $5$ are $(1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4)$ and $(6, 6).$ Hence, the number of favourable outcomes is $11.$
$\therefore \text{ P(A)}=\frac{\text{Number of favourable outcomesTotal}}{\text{ number of outcomes}}=\frac{11}{36}$
$\therefore\text{P}({\bar{\text{A}}})=1−\text{P}\text{(A)}=1−\frac{11}{36}=\frac{25}{36}$
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Question 142 Marks
In a simultaneous throw of a pair of dice, find the probability of getting: neither $9$ nor $11$ as the sum of the numbers on the faces.
Answer
When a pair of dice is thrown simultaneously, the sample space will be as follows: $S = \{(1, 1), (1, 2), (1, 3), (1, 4),$ ______ $(6, 5), (6, 6)\}$ Hence, the total number of outcomes is $36.$ Let A be the event of getting pairs whose sum is $9$ or $11.$ The pairs whose sum is $9$ are $(3, 6), (4, 5), (5, 4)$ and $(6, 3)$.And, the pairs whose sum is $11$ are $(5, 6)$ and $(6, 5).$ Hence, the number of favourable outcomes is $6.$
$\therefore \text{ P(A)}=\frac{\text{Number of favourable outcomesTotal}}{\text{ number of outcomes}}=\frac{6}{36}=\frac{1}{6}$
$\therefore P($sum of the pairs with neither $9$ nor $11) = 1 - P($sum of the pairs having $9$ or $1)$
$=1-\frac{1}{6}=\frac{5}{6}$
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Question 152 Marks
The probability that it will rain tomorrow is $0.85.$ What is the probability that it will not rain tomorrow$?$
Answer
Let $A$ be the event of raining tomorrow. The probability that it will rain tomorrow, $P(A),$ is $0.85.$ Since the event of raining tomorrow and not raining tomorrow are complementary to each other, the probability of not raining. $\text{P}(\bar{\text{A}}) = 1 − \text{P(A)} = 1 − 0.85 = 0.15$
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Question 162 Marks
A bag contains $6$ red, $8$ black and $4$ white balls. $A$ ball is drawn at random. What is the probability that ball drawn is not black$?$
Answer
Number of red balls $= 6$
Number of black balls $= 8$
Numberof white balls $= 4$
Total number of balls $= 6 + 8 + 4 = 18$
$\therefore$ Total number of cases $= 18$ Again,
number of balls that are not black $= 18 - 8 = 10$
Thus, the number of favourable cases is $10.$
$\therefore$ (the drawn ball is not black) $=\frac{\text{Number of favourable casesTotal}}{\text{number cases}}=\frac{10}{18}=\frac{5}{9}$
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Question 182 Marks
In a simultaneous throw of a pair of dice, find the probability of getting: a doublet.
Answer
When a pair of dice is thrown simultaneously, the sample space will be as follows:$S = \{(1, 1), (1, 2), (1, 3), (1, 4),$ ______ $(6, 5), (6, 6)\}$
Hence, the total number of outcomes is $36.$
Let $A$ be the event of getting doublets in the sample space.
The doublets in the sample space are $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5)$ and $(6, 6).$
Hence, the number of favourable outcomes is $6.$
$∴\text{ P(A)}=\frac{\text{Number of favourable outcomesTotal}}{\text{ number of outcomes}}=\frac{6}{36}=\frac{1}{6}$
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Question 192 Marks
In a simultaneous throw of a pair of dice, find the probability of getting: $8$ as the sum.
Answer
When a pair of dice is thrown simultaneously, the sample space will be as follows: $S = \{(1, 1), (1, 2), (1, 3), (1,4),$ ______ $(6, 5), (6, 6)\}$ Hence, the total number of outcomes is $36.$ Let $A$ be the event of getting pairs whose sum is $8$.Now, the pairs whose sum is $8$ are $(2, 6), (3, 5), (4, 4), (5, 3)$ and $(6, 2).$ Therefore, the total number of favourable outcomes is $5.$
$\therefore \text{ P(A)}=\frac{\text{Number of favourable outcomesTotal}}{\text{ number of outcomes}}=\frac{5}{36}$
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Question 202 Marks
Three coins are tossed together. Find the probability of getting: exactly two heads.
Answer
When $3$ coins are tossed together, the outcomes are as follows: $S = \{(h,h,h), (h, h, t), (h, t, h), (h, t, t), (t, h, h), (t, h, t), (t, t, h), (t, t, t)\}$Therefore, the total number of outcomes is $8.$ Let A be the event of getting triplets having exactly $2$ heads. Triplets having exactly 2 heads: $(h, h, t), (h, t, h), (t, h, h)$ Therefore, the total number of favourable outcomes is $3.$ $\text{P(A)}=\frac{\text{Number of favourable outcomesTotal}}{\text{ number of outcomes}}=\frac{3}{8}$
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Question 212 Marks
In a simultaneous throw of a pair of dice, find the probability of getting:an even number on one and a multiple of $3$ on the other.
Answer
When a pair of dice is thrown simultaneously, the sample space will be as follows:
$S = \{(1, 1), (1, 2), (1, 3), (1, 4),$ ______ $(6, 5), (6, 6)\}$ Hence, the total number of outcomes is $36.$
Let A be the event of getting pairs with an even number on one die and a multiple of $3$ on the other.
The pairs with an even number on one die and a multiple of $3$ on the other are $(2, 3), (2, 6), (4, 3), (4, 6), (6, 3)$ and $(6, 6)$.
Hence, the number of favourable outcomes is $6.$
$\therefore \text{ P(A)}=\frac{\text{Number of favourable outcomesTotal}}{\text{ number of outcomes}}=\frac{6}{36}=\frac{1}{6}$
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