[3 marks sum] - MATHS STD 8 Questions - Vidyadip

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[3 marks sum]

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15 questions · timed · auto-graded

Question 13 Marks

A bag contains $3$ white, $5$ black, and $2$ red balls, all of the same shape and size. A ball is drawn from the bag without looking into it, find the probability that the ball is drawn is:a white ball.

Answer

In a bag, $3$ balls are white
$2$ balls are red
$5$ balls are black
Total number of balls $=3+2+5=10$
Number of possible outcomes of white ball $=10$
and the number of favourable outcomes $=3$
$\therefore P(E)=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcome }}$
$=\frac{3}{10}$

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Question 23 Marks

A bag contains $3$ white, $5$ black, and $2$ red balls, all of the same shape and size.A ball is drawn from the bag without looking into it, find the probability that the ball is drawn is:a red ball.

Answer

In a bag, $3$ balls are white
$2$ balls are red
$5$ balls are black
Total number of balls $=3+2+5=10$
Number of possible outcomes of one red ball $=10$
and the number of favourable outcomes $=2$
$\therefore P(E)=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcome }}$
$ =\frac{2}{10}=\frac{1}{5}$

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Question 33 Marks

A letter is chosen from the word $\text{‘PENCIL’}$ what is the probability that the letter chosen is a consonant?

Answer

Total no. of letters in the word '$\text{PENCIL} =6$
Total Number of Consonant $= \text{'PNCL'}$ i.e. $4$
$P(E)=\frac{\text { Total No. of consonants }}{\text { Total No. of letters in the word PENCIL }}$
$ =\frac{4}{6}=\frac{2}{3}$

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Question 43 Marks

A coin is tossed twice. Find the probability of getting: two heads

Answer

Two heads
Possible number of favourable outcomes $=1 ($i.e. $\text{HH})$
Total number of possible outcomes $=4$
$\therefore P(E)=\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}$
$\therefore P(E)=\frac{1}{4}$

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Question 53 Marks

A coin is tossed twice. Find the probability of getting: two tails

Answer

Two tails
Possible number of favourable outcomes $=1 ($i.e. $\text{TT})$
Total number of possible outcomes $=4$
$\therefore P(E)=\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}$
$ =\frac{1}{4}$

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Question 63 Marks

A coin is tossed twice. Find the probability of getting: exactly one tail

Answer

Exactly one tail
Possible number of favourable outcomes $=2 ($i.e. $\text{TH}$ and $\text{HT} )$
Total number of possible outcomes $=4$
$\therefore P(E)=\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}$
$=\frac{2}{4}=\frac{1}{2}$

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Question 73 Marks

A coin is tossed twice. Find the probability of getting: exactly one head

Answer

Exactly one head
Possible number of favourable outcomes $=2 ($i.e. $\text{TH}$ and $\text{HT})$
Total number of possible outcomes $=4$
$\therefore P(E)=\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}$
$=\frac{2}{4}=\frac{1}{2}$

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Question 83 Marks

A coin is tossed. What is the probability of getting:a head?

Answer

On tossing a coin once,
Number of possible outcome $=2$
a head
Similarly, Favourable outcome of getting a head $=1$
But the number of possible outcomes $=2$
$\therefore P(E)=\frac{\text { Number of Favourable outcome }}{\text { Number of all possible outcome }}$
$=\frac{1}{2}$

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Question 93 Marks

A coin is tossed. What is the probability of getting:a tail?

Answer

On tossing a coin once,
Number of possible outcome $=2$
Favourable outcome getting a tail $=1$
$\Rightarrow$ number of Favourable outcome $=2$
$\therefore P(E)=\frac{\text { Number of Favourable outcome }}{\text { Number of all possible outcome }}$
$=\frac{1}{2}$

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Question 103 Marks

From $10$ identical cards, numbered $1, 2, 3, ……, 10$, one card is drawn at random. Find the probability that the number on the card drawn is a multiple of $2$

Answer

Total outcomes $=10$
i.e. $1,2,3,4,5,6,7,8,9,10$
Favourable outcomes $=5$
i.e. $2,4,6,8,10$
$P(E)=\frac{5}{10}=\frac{1}{2}$

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Question 113 Marks

A book contains $92$ pages. A page is chosen at random. What is the probability that the sum of the digits in the page number is $9$?

Answer

Number of pages of the book $=92$
Which are from $1$ to $92$
Number of possible outcomes $=92$
$\therefore$ Number of pages whose sum of its page is $9=10$
i.e. $9,18,27,36,45,54,63,72,81,90$
$\therefore P ( E )=\frac{10}{92}=\frac{5}{46} C$

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Question 123 Marks

A die is thrown, find the probability of getting:a number not greater than $4$.

Answer

A die has six numbers: $1,2,3,4,5,6$
$\therefore$ Number of possible outcomes $=6$
Number of favourable outcomes $=$ not greater than $4$ or numbers will be $1,2,3,4$ which are $4$ in numbers
$P(E)=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}$
$=\frac{4}{6}=\frac{2}{3}$

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Question 133 Marks

A die is thrown, find the probability of getting: a number greater than $4$

Answer

A die has six numbers: $1,2,3,4,5,6$
$\therefore$ Number of possible outcomes $=6$
a number greater than $4$
Numbers of favourable outcomes $=$ greater than four i.e. two number $5$ and $6$
$P(E)=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}$
$=\frac{2}{6}=\frac{1}{3}$

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Question 143 Marks

A bag contains six identical black balls. A child withdraws one ball from the bag without looking into it. What is the probability that he takes out: $(i)$ a white ball?$(ii)$ a black ball?

Answer

$\because$ There are $6$ black balls in a bag.
$\therefore$ number of possible outcomes $=6$
$(i)$ A white ball
As there is no white ball in the bag
$\therefore$ Its probability is zero $(0)=$ or $P(E)=0$
$(ii)$ a black ball
$\therefore$ Number of favourable outcomes $=1$
$\therefore P(E)=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}$
$=\frac{1}{6}$

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Question 153 Marks

A die is thrown, find the probability of getting:a prime number

Answer

A die has six numbers: $1,2,3,4,5,6$
$\therefore$ Number of possible outcomes $=6$
a prime number
Numbers of favourable outcomes $=$ a prime
number $=1,3,5$ which are $3$ in numbers
$P(E)=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}$
$=\frac{3}{6}=\frac{1}{2}$

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[3 marks sum] - MATHS STD 8 Questions - Vidyadip