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MATHS [4 marks sum]

Probability · ICSE STD 8 (ICSE - English Medium). 12 questions.

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[4 marks sum]

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12 questions · timed · auto-graded

Question 14 Marks
In a single throw of a die, find the probability of getting : $(i)\ 8\ (ii)\ $a number greater than $8\ (iii)$ a number less than $8$
Answer
On a die, the numbers are $1,2,3,4,5,6$ i.e. six.
$\therefore$ Number of possible outcome $=6$
$(i)$Number of a favourable outcomes $=0 (\because 8$ is not possible$)$
$\therefore P(E)=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcome }}=\frac{0}{6}=0$
$(ii)$Number greater than $8$ will be $0$
$\therefore P(E)=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcome }}=\frac{0}{6}=0$
$(iii)$Number less than $8$ will be $1,2,3,4,5,6$
$\therefore P(E)=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcome }}=\frac{6}{6}=1$
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Question 24 Marks
In a single throw of a die, find the probability that the number : $(i)$ will be an even number.$(ii)$ will be an odd number.$(iii)$ will not be an even number.
Answer
A die has six numbers: $1,2,3,4,5,6$
$\therefore$ Number of possible outcome $=6$
$(i)$ Number of favourable outcomes $=$ an even number i.e. $2, 4, 6$ which are $3$ in numbers
$\therefore P(E)=\frac{\text { Number of favourable outcome }}{\text { Number of all possible outcome }}$
$ =\frac{3}{6}=\frac{1}{2}$
$(ii) \ (iii)$ Number of favourable outcome $=$ not an even number i.e. odd numbers : $1,3,5$ which are $3$ in numbers
$\therefore P(E)=\frac{\text { Number of favourable outcome }}{\text { Number of all possible outcome }}$
$=\frac{3}{6}=\frac{1}{2}$
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Question 34 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 : not a black ball.
Answer
In a bag, $3$ balls are white
$2$ balls are red
$5$ balls are black
Total number of balls not a black ball $=3+2+5=10$
Number of possible outcomes $=10$
number of favourable outcomes not a black ball $=3+2=5$
$\therefore P(E)=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcome }}$
$=\frac{5}{10}=\frac{1}{2}$
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Question 44 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:not 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 $=10$
number of favourable outcomes $=3+5=8$
not a red ball
$\therefore P(E)=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcome }}$
$ =\frac{8}{10}=\frac{4}{5}$
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Question 54 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 black 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 black ball $=10$
and the number of favourable outcomes of one black ball $=5$
$\therefore P(E)=\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcome }}$
$=\frac{5}{10}=\frac{1}{2}$
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Question 64 Marks
In a single throw of a die, find the probability of getting a number $(i)$ greater than $2\ (ii)$ less than or equal to $2\ (iii)$ not greater than $2$.
Answer
A die has six numbers $=1,2,3,4,5,6$
$\therefore$ Number of possible outcomes $=6$
$\therefore P(E)=\frac{4}{6}=\frac{2}{3}$
$(ii)$Less than or equal to $2$
Number of favourable outcomes $=1,2$
$\therefore P(E)=\frac{2}{6}=\frac{1}{3}$
$(iii)$Not greater than $2$
Number of favourable outcomes $=1,2$
$\therefore P(E)=\frac{2}{6}=\frac{1}{3}$
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Question 74 Marks
A bag contains a black ball, a red ball and a green ball, all the balls are identical in shape and size. A ball is drawn from the bag without looking into it. What is the probability that the ball drawn is : $(i)$ a red ball $(ii)$ not a red ball $(iii)$ a white ball.
Answer
Total number of possible outcomes $=3$
$\therefore P(E)=\frac{1}{3}$
$(ii)$Not a red ball
Number of favourable outcomes
$=$ Green ball + Black ball
$=1+1+2$
$\therefore P(E)=\frac{2}{3}$
$(iii)$A white ball
Number of favourable outcomes $=0$
$\therefore P(E)=\frac{0}{3}=0$
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Question 84 Marks
Two dice are thrown at the same time. Find the probability that the sum of the two numbers appearing on the top of the dice is : less than $12$
Answer
Total outcomes $= 36$ i.e.
$(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)$
$(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)$
$(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)$
$(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)$
$(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)$
$(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)$
Favourable outcomes $=35 [$Except $(6,6)]$
$P(E)=\frac{35}{36}$
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Question 94 Marks
Two dice are thrown at the same time. Find the probability that the sum of the two numbers appearing on the top of the dice is : $12$
Answer
Total outcomes $=36$ i.e.
$(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)$
$(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)$
$(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)$
$(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)$
$(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)$
$(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)$
Favourable outcomes $=1$
i.e. $(6,6)$
$P(E)=\frac{1}{36}$
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Question 104 Marks
Two dice are thrown at the same time. Find the probability that the sum of the two numbers appearing on the top of the dice is : $0$
Answer
Total outcomes $= 36$ i.e.
$(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)$
$(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)$
$(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)$
$(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)$
$(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)$
$(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)$
Favourable outcomes $= 0$
$P(E) = \frac{0}{36}=0$
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Question 114 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 $2$ or $3$
Answer
Total outcomes $=10$
i.e. $1,2,3,4,5,6,7,8,9,10$
Favourable outcomes $=7$
i.e. $2,3,4,6,8,9,10$
$P(E)=\frac{7}{10}$
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Question 124 Marks
Two coins are tossed together. What is the probability of getting: $(i)$ at least one head $(ii)$ both heads or both tails.
Answer
$\because$ A coin has two faces Head and Tail or ${H_1 T}$
$\therefore$ Two coins are tossed
$\therefore$ Number of coins $=2 \times 2=4$
which are $\text{HH , HT , TH , TT}$
$(i)$At least one head, then
Number of outcomes $=3$
$\therefore P(E)=\frac{\text { Number of favourable outcome }}{\text { Number of all possible outcome }}$
$ =\frac{3}{4}$
$(ii)$When both head or both tails, then
Number of outcomes $=2$
$\therefore P(E)=\frac{\text { Number of favourable outcome }}{\text { Number of all possible outcome }}$
$\frac{2}{4}=\frac{1}{2}$
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[4 marks sum] - MATHS STD 8 Questions - Vidyadip