Chapter 3:Pair of Linear Equations in Two Variables

Chapter 5: Arithmetic Progression

Chapter 8: Introduction to Trigonometry

Chapter 9:Some Applications of Trigonometry

Chapter 12:Area Related to Circles

Probability

(i) Probability of an event E + Probability of the event ‘not E’ = ___________.

(ii) The probability of an event that cannot happen is __________. Such an event is called ________.

(iii) The probability of an event that is certain to happen is _________. Such an event is called _________.

(iv) The sum of the probabilities of all the elementary events of an experiment is __________.

(v) The probability of an event is greater than or equal to and less than or equal to __________.

(i) Probability of an event E + Probability of the event ‘not E’ = **1**

(ii) The probability of an event that cannot happen is **0** Such an event is called an ** impossible event **

(iii) The probability of an event that is certain to happen is **1**. Such an event is called **a sure or certain event**

(iv) The sum of the probabilities of all the elementary events of an experiment is **1**

(v) The probability of an event is greater than or equal to 0 and less than or equal to **1**

(i) A driver attempts to start a car. The car starts or does not start.

(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.

(iii) A trial is made to Solution: a true-false question. The Solution: is right or wrong.

(iv) A baby is born. It is a boy or a girl.

(i) It is not have equally likely outcomes as the car may or may not start depending upon various factors like fuel or whether car state or not, etc.

(ii) this statement does not have equally likely outcomes as it depends the player's ability or player may shoot or miss the shot.

(iii) This statement has equally likely.

(iv) This statement also has equally likely

Tossing of a coin is a fair way of deciding because the number of possible outcomes are only 2 head or tail.So tossing is unpredictable and is considered to be completely fair.

(A) 2/3 (B) -1.5 (C) 15% (D) 0.7

the above option (B) -1.5 cannot be the probability of an event.As probability lie between 1 and 0

Here We know that,

P(E)+P(not E) = 1

P(E) = 0.05

P(not E) = 1-P(E)

P(not E) = 1-0.05

∴ P(not E) = 0.95

(i) an orange flavored candy?

(ii) a lemon flavored candy?

i) Number of orange flavored candies in the = 0

∴ The probability of orange flavored candies = 0/1 = 0

(ii) As only lemon flavored candies in the bag , P(lemon flavored candies) = 1 (or 100%)

Given, P(E) = 0.992

We know that

P(E)+P(not E) = 1

Or, P(not E) = 1–0.992 = 0.008

∴ The probability that the 2 students have the same birthday is 0.008

(i) red?

(ii) not red?

the total Number of balls = 5+3 = 8

P(E) =
Number of favourable outcomes
/
Total number of outcomes

(i) Probability of drawing red balls = P (red balls) =

no. of red balls
/
total no. of balls

=
3
/
8

(ii) Probability of drawing black balls = P (black balls) =

no. of black balls
/
total no. of balls

=
5
/
8

(i) red?

(ii) white?

(iii) not green?

The Total no. of balls = 5+8+4 = 17

P(E) =
Number of favourable outcomes
/
Total number of outcomes

(i) Total number of red balls = 5

P (red ball) =
5
/
17

= 0.29
(ii) Total number of white balls = 8

P (white ball) =
8
/
17

= 0.47(iii) Total number of green balls = 4

P (green ball) =

4
/
17

= 0.23∴ P (not green) = 1-P(green ball) = 1-(

4
/
17

) = 0.77

(i) will be a 50 p coin?

(ii) will not be a ₹5 coin?

Total no. of coins = 100+50+20+10 = 180

(i) Total number of 50 p coin = 100

P (50 p coin) = 100/180 = 5/9 = 0.55

(ii) Total number of ₹5 coin = 10

P (₹5 coin) = 10/180 = 1/18 = 0.055

∴ P (not ₹5 coin) = 1-P (₹5 coin) = 1-0.055 = 0.945

The total number of fish in the tank = 5+8 = 13

Total number of male fish = 5

P (male fish) = 5/13 = 0.38

(i) 8?

(ii) an odd number?

(iii) a number greater than 2?

(iv) a number less than 9?

Total number of possible outcomes = 8

(i) Total number of favourable events (i.e. 8) = 1

∴ P (of 8) = 1/8 = 0.125

(ii) Total number of odd numbers = 4 (1, 3, 5 and 7)

P (pointing at an odd number) = 4/8 = 1/2 = 0.5

(iii) Total numbers greater than 2 = 6 (3, 4, 5, 6, 7 and 8)

P (pointing at a number greater than 4) = 6/8 = 3/4 = 0.75

(iv) Total numbers less than 9 = 8 (1, 2, 3, 4, 5, 6, 7, and 8)

P (pointing at a number less than 9) = 8/8 = 1

(i) a prime number;

(ii) a number lying between 2 and 6;

(iii) an odd number.

Total possible number when a dice is thrown = 6 (1, 2, 3, 4, 5, and 6)

i) Total number of prime numbers = 3 (2, 3 and 5)

P (getting a prime number) = 3/6 = 1/2 = 0.5

(ii) Total numbers lying between 2 and 6 = 3 (3, 4 and 5)

P (getting a number between 2 and 6) = 3/6 = 1/2 = 0.5

(iii) Total number of odd numbers = 3 (1, 3 and 5)

P (getting an odd number) = 3/6 = 1/2 = 0.5

(i) a king of red colour

(ii) a face card

(iii) a red face card

(iv) the jack of hearts

(v) a spade

(vi) the queen of diamonds

Total number of possible outcomes = 52

(i) Total numbers of king of red colour = 2

P (king of red colour) = 2/52 = 1/26 = 0.038

(ii) Total numbers of face cards = 12

P (getting a face card) = 12/52 = 3/13 = 0.23

(iii) Total numbers of red face cards = 6

P ( red face cards ) = 6/52 = 3/26 = 0.11

(iv) Total numbers of jack of hearts = 1

P (getting a jack of hearts) = 1/52 = 0.019

(v) Total numbers of king of spade = 13

P (getting a king of spade ) = 13/52 = 1/4 = 0.25

(vi) Total numbers of queen of diamonds = 1

P (getting a queen of diamonds) = 1/52 = 0.019

(i) What is the probability that the card is the queen?

(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?

Total numbers of cards = 5

(i) Numbers of queen = 1

P (picking a queen) = 1/5 = 0.2

(ii) the total numbers of cards left after drawn queen is = 4

(a) Total numbers of ace = 1

P (picking an ace) = 1/4 = 0.25

(b) Total numbers of queen = 0

P (picking a queen) = 0/4 = 0

Numbers of pens = defective pens + good pens

∴ Total number of pens = 132+12 = 144 pens

P(picking a good pen) = 132/144 = 11/12 = 0.916

(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

(i)Numbers of defective bulbs = 4

The total numbers of bulbs = 20

∴ Probability of getting a defective bulb = P (defective bulb) = 4/20 = 1/5= 0.2

(ii) the total numbers of bulbs left when 1 non-defective bulb is drawn is 9

So, the total numbers of events (or outcomes) = 19

Numbers of defective bulbs = 19-4 = 15

So, the probability that the bulb is not defective = 15/19 = 0.789

(i) a two-digit number

(ii) a perfect square number

(iii) a number divisible by 5.

The total numbers of discs = 90

(i) Total number of discs having two digit numbers = 81

P (bearing a two-digit number) = 81/90 = 9/10 = 0.9

(ii) perfect square numbers(1, 4, 9, 16, 25, 36, 49, 64 and 81) = 9

P (getting a perfect square number) = 9/90 = 1/10 = 0.1

(iii) Total numbers which are divisible by 5 = 18 (5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85 and 90)

P (getting a number divisible by 5) = 18/90 = 1/5 = 0.2

The die is thrown once. What is the probability of getting

(i) A?

(ii) D?

The total number of possible events = 6

(i) The total number of faces having A = 2

P (getting A) = 2/6 = 1/3 = 0.33

(ii) The total number of faces having D = 1

P (getting D) = 1/6 = 0.166

Lenght= 3m

Breadth =2m

So, the area of the rectangle = (3×2) m^{2} = 6 m^{2}

and,

The area of the circle = πr^{2} = π(½)^{2} m^{2} = π/4 m^{2} = 0.78

∴ The probability that die will land inside the circle = [(π/4)/6] = π/24 or, 0.78/6 = 0.13

(i) She will buy it?

(ii) She will not buy it?

The total numbers of outcomes i.e. pens = 144

Given, numbers of defective pens = 20

∴ The numbers of non defective pens = 144-20 = 124

(i) P (buying) = 124/144 = 31/36 = 0.86

(ii) she will not buy them = 20

So, P (not buying) = 20/144 = 5/36 = 0.138

(ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability 1/11. Do you agree with this argument? Justify your Solution:.

If 2 dices are thrown, the possible events 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), (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)

So, the total events: 6×6 = 36

(i)P(sum of 2) = 1/36

E (sum 3) = (1,2) and (2,1)

P(sum 3) = 2/36

Similarly,

E (sum 4) = (1,3), (3,1), and (2,2)

P (sum 4) = 3/36

E (sum 5) = (1,4), (4,1), (2,3), and (3,2)

P (sum 5) = 4/36

E (sum 6) = (1,5), (5,1), (2,4), (4,2), and (3,3)

P (sum 6) = 5/36

E (sum 7) = (1,6), (6,1), (5,2), (2,5), (4,3), and (3,4)

P (sum 7) = 6/36

E (sum 8) = (2,6), (6,2), (3,5), (5,3), and (4,4)

P (sum 8) = 5/36

E (sum 9) = (3,6), (6,3), (4,5), and (5,4)

P (sum 9) = 4/36

E (sum 10) = (4,6), (6,4), and (5,5)

P (sum 10) = 3/36

E (sum 11) = (5,6), and (6,5)

P (sum 11) = 2/36

E (sum 12) = (6,6)

So, P (sum 12) = 1/36

Event: Sum on 2 dice |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |

Probability |
1/36 |
2/36 |
3/36 |
4/36 |
5/36 |
6/36 |
5/36 |
4/36 |
3/36 |
2/36 |
1/36 |

(ii) The argument is not correct as the possible outcomes is 36 and not 11.

The total number of outcomes = 8 (HHH, HHT, HTH, THH, TTH, HTT, THT, TTT)

Total outcomes in which Hanif will lose the game = 6 (HHT, HTH, THH, TTH, HTT, THT)

P (losing the game) = 6/8 = 3/4= 0.75

(i) 5 will not come up either time?

(ii) 5 will come up at least once?

[Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]

Outcomes 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), (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)

So, the total number of outcome = 6×6 = 36

(i) **Method 1:**

P(5 comes in first throw) = 6/36,

P( comes in second throw) = 6/36 and

P(not B) = 5/6

So, P(not A) = 1-(6/36) = 5/6

∴ The required probability = (5/6)×(5/6) = 25/36

(ii) Number of events when 5 comes at least once = 11

∴ The r probability = 11/36

(i) If two coins are tossed simultaneously there are three possible outcomes—two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 1/3

(ii) If a die is thrown, there are two possible outcomes—an odd number or an even number. Therefore, the probability of getting an odd number is 1/2

(i) All the possible events are (H,H); (H,T); (T,H) and (T,T)

So, P ( heads) = 1/4= 0.25

and, P (getting one of the each) = 2/4 = 1/2 = 0.5

∴ This statement is incorrect.

(ii)Since the two outcomes are equally likely,

this statement is correct.