[4 marks sum]

Question 14 Marks

A bag contains twenty Rs 5 coins, fifty Rs 2 coins and thirty Re 1 coins. If it is equally likely that one of the coins will fall down when the bag is turned upside down, what is the probability that the coin : will neither be a Rs 5 coin nor be a Re 1 coin?

Answer

Total number of coins =20+50+30=100
Total possible outcomes=100=n(s)
Number of favorable outcomes for niether Re1 nor Rs5 coins= Number of favourable outcomes for Rs2 coins=50= n(E)
Probability (neither Re 1 nor Rs 5 coin)=
$\frac{n(E)}{n(s)}=\frac{50}{100}=\frac{1}{2}$

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Question 24 Marks

A game of numbers has cards marked with 11, 12, 13, … 40. A card is drawn at random. Find the probability that the number on the card drawn is : 1) A perfect square 2) Divisible by 7

Answer

Total number of outcomes $=30$
1) The perfect squares lying between 11 and 40 are 16,25 and 36 .
So the number of possible outcomes 3
$\therefore$ Probability that the number on the card drawn is a perfect square
$
=\frac{\text { Number of possible outcomes }}{\text { Total number of outcomes }}=\frac{3}{30}=\frac{1}{10}
$
2) The numbers from 11 to 40 that are divisible by 7 are $14,21,28$ and 35 .
So the number of possible outcomes
The probability that the number on the card drawn is divisible by 7
$=\frac{\text { Number of possible outcomes }}{\text { Total number of outcomes }}=\frac{4}{30}=\frac{2}{15}$

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Question 34 Marks

A bag contains $5$ white balls, $6$ red balls and $9$ green balls. A ball is drawn at random from the bag. Find the probability that the ball is drawn is: $1)$ a green ball $2)$ a white or a red ball $3)$ is neither a green ball nor a white ball.

Answer

Total number of balls $= 5 + 6 + 9 = 20$
$1)$ Number of green balls $9$ Number of favourable cases
$\therefore P($Green ball$) =\frac{\text { Number of favourable cases }}{\text { Total number of balls }}=\frac{9}{20}$
$2)$ Number of white balls $5,$ Number of red balls $6$
Number of favourable cases $= 5 + 6 = 11$
$\therefore P($White ball or Red ball$) =\frac{\text { Number of favourable cases }}{\text { Total number of balls }}=\frac{11}{20}$
$3) P($Neither green ball nor white ball$) P($Red ball$)$
$=\frac{\text { Number of Red balls }}{\text { Total number of balls }}=\frac{6}{20}=\frac{3}{10}$

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Question 44 Marks

Two dice are thrown simultaneously. What is the probability that: 4 will not come up either time? 4 will come up at least once?

Answer

When two dice are thrown , total possible outccomes=36
Favorable outcomes for 4 will not come up either time:
{(1,1)(1,2)(1,3)(1,5)(1,6)}
{(2,1)(2,2)(2,3)(2,5)(2,6)}
{(3,1)(3,2)(3,3)(3,5)(3,6)}
{(5,1)(5,2)(5,3)(5,5)(5,6)}
{(6,1)(6,2)(6,3)(6,5)(6,6)}
Number of favorable outcomes = 25
p(4 will not come up) = 25/36
P(4 will come up ones) = 1- p(4 will not come up either time)
p(4 will come up once) =$1-\frac{25}{36}$
p(4 will come up once) = (36-25)/36=11/36

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Question 54 Marks

A circle with diameter 20 cm is drawn somewhere on a rectangular piece of paper with length 40 cm and width 30 cm. This paper is kept horizontal on table top and a die, very small in size, is dropped on the rectangular paper without seeing towards it. If the die falls and lands on paper only, find the probability that it will fall and land: inside the circle

Answer

Diameter of the circle= 20
Radius= 10 cm
Area of circle $=\pi r^2=\frac{22}{7} \times 10 \times 10 \times=\frac{2200}{7} cm ^2$
Length of paper =40 cm
Width of paper =30 cm
Area of paper $=1200 cm ^2$
|Total possible outccomes=arrea of rectangular paper
Since paper is kept on table top and die falls and land on paper.
Number of favorable outcomes= are of circle.
P(inside thhe circle)=$\frac{\text { Area of arde }}{\text { Area of rectangular paper }}=\frac{\frac{2200}{7}}{1200}=\frac{11}{42}$

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Question 64 Marks

Two dice are rolled together. Find the probability of getting: a total of at least 10

Answer

n(s) = 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)
In throwing a dice, total possible outcomes={1,2,3,4,5,6}
n(s)=6
for two dice, n(s)= 6 x 6=36
E= event of getting a total of at least 10={(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)}
n(E)=6
Probability of getting a total of at least 10=
$\frac{n(E)}{n(s)}=\frac{6}{36}=\frac{1}{6}$

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Question 74 Marks

In a single throw of two dice, find the probability of: a total of at most 10

Answer

The number of possible outcomes is $6 \times 6=36$
We write them as given below:
1,11,21,,31,41,51,6
2,12,22,32,42,52,6
3,13,23,33,43,53,6
4,14,24,34,44,54,6
5,15,25,35,45,55,6
6,16,26,36,46,56,6
n(s)=36
E= Event of getting a total of at most 10=
{(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)}
{(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)}
{(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)}
{(5,1)(5,2)(5,3)(5,4)(5,5)}
{(6,1)(6,2)(6,3)(6,4)}
Thereefore total number of favorable ways=33=n(E)
Probability of getting a total of at most 10=
$\frac{n(E)}{n(s)}=\frac{33}{36}=\frac{11}{12}$

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Question 84 Marks

In a single throw of two dice, find the probability of: a number less than 3 on each dice

Answer

The number of possible outcomes is$6 \times 6=36$,We write them as given below:
1,11,21,,31,41,51,6
2,12,22,32,42,52,6
3,13,23,33,43,53,6
4,14,24,34,44,54,6
5,15,25,35,45,55,6
6,16,26,36,46,56,6
n(s)=36
E= Event of getting a doublet={(1,1)(2,2)(3,3)(4,4)(5,5)(6,6)}
n(E)=6
Probability of getting a doublet =$\frac{n(E)}{n(S)}=\frac{6}{36}=\frac{1}{6}$

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Mathematics [4 marks sum]

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