Lesson: Probability

Exercise 15.1(25)

Question: 1

Complete the following statements:

(i) Probability of an event Probability of the

E

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______.

Solution

(i)

1

(ii) , impossible event

0

(iii) , certain event

1

(iv)

1

(v)

0,1

Question: 2

Which of the following experiments have equally

likely outcomes? Explain.

(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 answer a true-false question.

The answer is right or wrong.

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

Solution

(i) The event will depend on several factors like

whether the driver will be able to start the car

or not. In both the situations, the factors are

not the same. Thus, the event does not have

equally likely outcomes.

(ii) In this event, the information on the player’s

ability is not given. And the event depends on

the player’s ability. Thus, the event does not

have equally likely outcomes.

(iii) In the event, either the answer will be correct or

not. Thus, there will be two possible outcomes

and event has equally likely outcomes.

(iv) In the event, either the baby will be boy or girl.

Thus, there will be two possible outcomes and

event has equally likely outcomes.

Question: 3

Why is tossing a coin considered to be a fair way of

deciding which team should get the ball at the

beginning of a football game?

Solution

In the event of tossing a coin, either it will give head

or tail. Thus, there are two possible outcomes of

tossing a coin that are equally likely outcomes. The

result of tossing a coin is completely unpredictable

that makes it a fair way of deciding the team that

will get the ball.

Question: 4

Which of the following cannot be the probability of

an event?

(A)

2

3

(B)

1.5

(C)

15%

(D)

0.7

Solution

(B)

We know that:

Probability of an event .

E,

P E 0

Probability of an event .

E,

P E 1

Thus, the probability of an event cannot be more

than or negative.

1

Hence, cannot be the probability of an event.

1.5

Question: 5

If , what is the probability of ‘not ’?

P E 0.05

E

Solution

Given:

P E 0.05

The formula for the probability of a non-occurring

event is

P not Eor E 1 P E

Substitute the value of occurring event in the

formula.

P not E 1 P E

1 0.05

0.95

Thus, the probability of ‘not ’ is 0.95.

E

Question: 6

A bag contains lemon flavored candies only. Malini

takes out one candy without looking into the bag.

What is the probability that she takes out

(i) an orange flavored candy?

(ii) a lemon flavored candy?

Solution

(i) It is given that Malini has only lemon flavored

candies in the bag. The bag does not carry

orange flavored candies. Thus, Malini will get

lemon flavored candies every time and the

event of taking out an orange flavored candy is

an impossible event.

Hence, the probability of taking out an orange

flavored candy is .

0

(ii) It is given that Malini has only lemon flavore

candies in the bag. Thus, Malini will get lemon

flavored candies every time and the event of

taking out a lemon flavoured candy is a sure

event.

Hence, the probability of taking out a lemon

flavored candy is .

1

Question: 7

It is given that in a group of 3 students, the

probability of 2 students not having the same

birthday is 0.992. What is the probability that the 2

students have the same birthday?

Solution

Given:

Probability of 2 students not having the same

birthday is .

0.992

That is,

P not Eor E 0.992

The formula for the probability of a non-occurring

event is

P not Eor E 1 P E

Substitute the value of occurring event in the

formula.

0.992 1 P E

P E 1 0.992

P E 0.008

Thus, the probability of 2 students having the same

birthday is .

0.008

Question: 8

A bag contains 3 red balls and 5 black balls. A ball is

drawn at random from the bag.

What is the probability that the ball drawn is

(i) red?

(ii) not red?

Solution

(i) Let be the event that the ball drawn is red.

E

Total number of balls

3 5 8

Number of red balls

3

Thus,

Number of redballs

P E

Total number of balls

3

8

Hence, the probability of getting red ball is

3

8

(ii) The formula for the probability of a non-

occurring event is

P not Eor E 1 P E

Substitute the value of occurring event in the

Formula.

3

P not Eor E 1

8

8 3

8

5

8

Hence, the probability of not getting red ball is

5

8

Question: 9

A box contains 5 red marbles, 8 white marbles and 4

green marbles. One marble is taken out of the box at

random. What is the probability that the marble

taken out will be

(i) red?

(ii) white?

(iii) not green?

Solution

Total number of marbles

5 8 4 17

(i) Let be the event that the marble taken out is

E

red.

Number of red marbles

5

Thus,

Number of redmarbles

P E

Total number of marbles

5

17

Hence, the probability of taking out red marble

is

5

17

(ii) Let be the event that the marble taken out is

E

white.

Number of white marbles

8

Thus,

Number of whitemarbles

P E

Total number of marbles

8

17

Hence, the probability of taking out white

marble is

8

17

(iii) Let be the event that the marble taken out is

E

green.

Number of green marbles

4

Thus,

Number of green marbles

P E

Total number of marbles

4

17

Hence, the probability of taking out green

marble is

4

17

The formula for the probability of a non-

occurring event is

P not Eor E 1 P E

Substitute the value of occurring event in the

formula.

4

P not Eor E 1

17

17 4

17

13

17

Hence, the probability of not taking out green

marble is .

13

17

Question: 10

A piggy bank contains hundred 50p coins, fifty Re 1

coins, twenty Rs 2 coins and ten Rs 5 coins. If it is

equally likely that one of the coins will fall out when

the bank is turned upside down, what is the

probability that the coin

(i) will be a 50p coin?

(ii) will not be a Rs 5 coin?

Solution

Total number of coins

100 50 20 10 180

(i) Let be the event that the coin fall out is 50p

E

coin.

Number of 50p coin

100

Thus,

50p coinNumber of

P E

Total number of coins

100

180

5

9

Hence, the probability of falling out 50p coin is

5

9

(ii) Let be the event that the coin fall out is Rs 5

E

coin.

Number of Rs 5 coin

10

Thus,

Number of

P E

Total number o

Rs5c

f c

oin

oins

10

180

1

18

Hence, the probability of falling out Rs 5 coin is

1

18

The formula for the probability of a non-

occurring event is

P not Eor E 1 P E

Substitute the value of occurring event in the

formula.

1

P not Eor E 1

18

18 1

18

17

18

Hence, the probability of not falling out Rs 5

coin is .

17

18

Question: 11

Gopi buys a fish from a shop for his aquarium. The

shopkeeper takes out one fish at random from a

tank containing 5 male fish and 8 female fish (see

Fig. 15.4). What is the probability that the fish taken

out is a male fish?

Fig. Exc_15.1_11

Solution

Total number of fishes

5 8 13

Let be the event that the fish taken out is a male

E

fish.

Number of male fishes

5

Thus,

Number of malefishes

P E

Total number of fishes

5

13

Hence, the probability of getting male fish is

5

13

Question: 12

A game of chance consists of spinning an arrow

which comes to rest pointing at one of the numbers

(see Fig. 15.5), and these are equally

1,

2,

3,

4,

5,

6,

7,

8

likely outcomes. What is the probability that it will

point at

(i) ?

8

(ii) an odd number?

(iii) a number greater than 2.

(iv) a number less than 9.

Fig. Exc_15.1_12

Solution

In the given figure,

Total possible outcomes

8

(i) Let be the event that the arrow will point at

E

8.

Number of favourable outcomes

1

Thus,

Number of

P E

Total number o

favourable o

f possible

utcomes

outcomes

1

8

Hence, the probability of arrow pointing at is

8

.

1

8

(ii) The figure contains odd numbers .

1,3,5,7

Thus, total odd numbers

4

Let be the event that the arrow will point at

E

an odd number.

Number of favourable outcomes

4

Thus,

Number of

P E

Total number o

favourable o

f possible

utcomes

outcomes

4

8

1

2

Hence, the probability of arrow pointing at an

odd number is .

1

2

(iii) The numbers in the figure that are greater than

2 are .

3,

4,

5,

6,

7,

8

Thus, total of such numbers

6

Let be the event that the arrow will point at a

E

number greater than 2.

Number of favourable outcomes

6

Thus,

Number of

P E

Total number o

favourable o

f possible

utcomes

outcomes

6

8

3

4

Hence, the probability of arrow pointing at a

number greater than 2 is .

3

4

(iv) The numbers in the figure that are less than 9

are .

1,

2,

3,

4,

5,

6,

7,

8

Thus, total of such numbers

8

Let be the event that the arrow will point at a

E

number less than 9.

Number of favourable outcomes

8

Thus,

Number of

P E

Total number o

favourable o

f possible

utcomes

outcomes

8

8

1

Hence, the probability of arrow pointing at a

number less than 9 is .

1

Question: 13

A die is thrown once. Find the probability of getting

(i) a prime number;

(ii) a number lying between 2 and 6;

(iii) an odd number;

Solution

On throwing a dice, the possible outcomes are

.

1,

2,

3,

4,

5,

6,

Thus, the total number of possible outcomes

6

(i) The dice has prime numbers and .

2,

3

5

Total prime numbers

3

Let be the event of having a prime number.

E

Number of favorable outcomes

3

Thus,

Number of

P E

Total number o

favourable o

f possible

utcomes

outcomes

3

6

1

2

Hence, the probability of having a prime

number is .

1

2

(ii) The dice has numbers and lying between

3,

4

5

and .

2

6

Total such numbers

3

(iii) Let be the event of having a number lying

E

between and .

2

6

Number of favourable outcomes

3

Thus,

Number of

P E

Total number o

favourable o

f possible

utcomes

outcomes

3

6

1

2

(iv) Hence, the probability of having a number lying

between and . is .

2

6

1

2

(v) The dice has odd numbers and .

1,

3

5

Total odd numbers

3

Let be the event of having an odd number.

E

Number of favourable outcomes

3

Thus,

Number of

P E

Total number o

favourable o

f possible

utcomes

outcomes

3

6

1

2

Hence, the probability of having an odd

number is .

1

2

Question: 14

One card is drawn from a well-shuffled deck of 52

cards. Find the probability of getting

(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

Solution

We know that a well-shuffled deck contains a total

number of 52 cards.

(i) Number of red kings in the deck

2

Let be the event of getting a red king.

E

Thus,

Number of redkings

P E

Total number of cardsin thedeck

2

52

1

26

Hence, the probability of getting a red king is

.

1

26

(ii) Number of face cards in the deck

12

Let be the event of getting a face card.

E

Thus,

Number of facecards

P E

Total number of cardsin thedeck

12

52

3

13

Hence, the probability of getting a face card is

.

3

13

(iii) Number of red face cards in the deck

6

Let be the event of getting a red face card.

E

Thus,

Number of red facecards

P E

Total number of cardsin thedeck

6

52

3

26

Hence, the probability of getting a red face card

is .

3

26

(iv) Number of jack of hearts in the deck

1

Let be the event of getting the jack of hearts.

E

Thus,

Number of jack of hearts

P E

Total number of cardsin thedeck

1

52

Hence, the probability of getting the jack of

hearts is .

1

52

(v) Number of spade cards in the deck

13

Let be the event of getting a spade.

E

Thus,

Number of spadecards

P E

Total number of cardsin thedeck

13

52

1

4

Hence, the probability of getting a spade is .

1

4

(vi) Number of queens of diamonds in the deck

1

Let be the event of getting the queen of

E

diamond.

Thus,

Number of queensof diamond

P E

Total number of cardsin thedeck

1

52

Hence, the probability of getting the queen of

diamond is .

1

52

Question: 15

Five cards--the ten, jack, queen, king and ace of

diamonds, are well-shuffled with their face

downwards. One card is then picked up at random.

(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?

Solution

(i) Given that the total number of cards is .

5

Number of queens

1

Let be the event of getting a queen.

E

Thus,

Number of queens

P E

Total number of cards

1

5

Hence, the probability of getting a queen is .

1

5

(ii) Total number of cards remaining after drawing

queen

4

(a) Number of aces

1

Let be the event of getting an ace.

E

Thus,

Number of aces

P E

Total number of cards

1

4

Hence, the probability of getting an ace is

.

1

4

(b) Number of queens after drawing queen

0

Let be the event of getting a queen.

E

Thus,

Number of queens

P E

Total number of cards

0

4

Hence, the probability of having a queen is

.

0

Question: 16

12 defective pens are accidentally mixed with 132

good ones. It is not possible to just look at a pen and

tell whether or not it is defective. One pen is taken

out at random from this lot. Determine the

probability that the pen taken out is a good one.

Solution

The total number of pens which include defective as

well as good ones

12 132 144

Number of good ones

132

Let be the event of taking out a good pen.

E

Thus,

Number of goodpens

P E

Total number of pens

132

144

11

12

Hence, the probability of getting a good pen is .

11

12

Question: 17

(i) A lot of 20 bulbs contain 4 defective ones. One

bulb is drawn at random from the lot. What is

the probability that this bulb is defective?

(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?

Solution

Given that the lot contains 20 bulbs.

(i) Number of defective bulbs in the lot

4

Let be the event of drawing a defective bulb.

E

Thus,

Number of defectivebulbs

P E

Total number of bulbs

4

20

1

5

Hence, the probability of drawing a defective

bulb is .

1

5

(ii) Total number of bulbs remaining after drawing

a non-defective bulb .

20 1 19

Number of non-defective bulbs

20 4 16

Number of non-defective bulbs remaining after

drawing a non-defective bulb .

16 1 15

Let be the event of drawing a non-defective

E

bulb.

Thus,

Number of non defectivebulbs

P E

Total number of bulbs

15

19

Hence, the probability of drawing a non-

defective bulb is .

15

19

Question: 18

A box contains 90 discs which are numbered from 1

to 90. If one disc is drawn at random from the box,

find the probability that it bears

(i) a two-digit number

(ii) a perfect square number

(iii) a number divisible by 5

Solution

Given that the box contains a total number of 90

discs.

(i) Number of discs with two-digit numbers

between and

1

90

81

Let be the event of drawing a disc of two-

E

digit number.

Thus,

Number of disc with two digit number

P E

Total number of discs

81

90

9

10

Hence, the probability of drawing a disc with a

two-digit number is .

9

10

(ii) Perfect squares between

and

1

90

1,

4,

9,

16,

25,

36,

49,

64,

81

Number of discs with perfect squares between

and

1

90

9

Let be the event of drawing a disc with a

E

perfect square number.

Thus,

Number of disc with perfect squarenumber

P E

Total number of discs

9

90

1

10

Hence, the probability of drawing a disc with a

perfect square number is .

1

10

(iii) Numbers divisible by 5 between and are

1

90

5,

10,

15,

20,

25,

30,

35,

40,

45,

50,

55,

60,

65,

70,

75,

80,

.

85,

90

Number of discs with a number divisible by 5

between and

1

90

18

Let be the event of drawing a disc with a

E

number divisible by 5.

Thus,

Number of disc with a number divisibleby 5

P E

Total number of discs

18

90

1

5

Hence, the probability of drawing a disc with a

number divisible by 5 is .

1

5

Question: 19

A child has a die whose six faces show the letters as

given below:

Fig. Exc15.1_19

The die is thrown once. What is the probability of

getting

(i) ?

A

(ii) ?

D

Solution

As shown in the given figure, the die has 6 faces.

(i) Number of faces with

A 2

Let be the event of getting face with .

E

A

Thus,

Number of faceswith A

P E

Total number of faces

2

6

1

3

Hence, the probability of getting the face with

is .

A

1

3

(ii) Number of faces with

D 1

Let be the event of getting face with .

E

D

Thus,

Number of faceswithD

P E

Total number of faces

1

6

Hence, the probability of getting the face with

is .

D

1

6

Question: 20

Suppose you drop a die at random on the

rectangular region shown in Fig. 15.6. What is the

probability that it will land inside the circle with

diameter 1m?

Fig. Exc15.1_20

Solution

Given that:

Length of the rectangle

3m

Breadth of the rectangle

2m

Area of the rectangle

length breadth

2

3 2 6m

Diameter of the circle

1m

Radius of the circle

1

r m

2

Area of the circle

2

πr

Area of the circle

2

1 π

π

2 4

Let be the event of falling the die inside the circle.

E

Thus,

Area of thecircle

P E

Area of therectangle

π

4

6

π

4 6

π

24

Hence, the probability of falling the die inside the

circle is .

π

24

Question: 21

A lot consists of 144 ball pens of which 20 are

defective and the others are good. Nuri will buy a

pen if it is good, but will not buy if it is defective.

The shopkeeper draws one pen at random and gives

it to her. What is the probability that

(i) She will buy it?

(ii) She will not buy it?

Solution

Total number of ball pens in the lot

144

Number of defective ball pens

20

Number of good ball pens

144 20 124

(i) Let be the event of drawing a good ball pen.

E

Thus,

Number of goodball pens

P E

Total number of ball pens

124

144

31

36

Hence, the probability that Nuri will buy the

ball pen is .

31

36

(ii)

31

P E

36

The formula for the probability of a non-

occurring event is

P not Eor E 1 P E

Substitute the value of occurring event in the

formula.

P not E 1 P E

31

1

36

36 31

36

5

36

Hence, the probability that Nuri will not buy

the ball pen is .

5

36

Question: 22

Refer to Example 13.

(i) Complete the following table:

Event:

‘Sum on 2

dice’

2

3

4

5

6

7

8

9

10

11

12

Probability

1

36

5

36

1

36

(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

answer.

Example 13: Two dice, one blue and one grey, are

thrown at the same time. Write down all the

possible outcomes. What is the probability that the

sum of the two numbers appearing on the top of the

dice is

Solution

(i) Number of possible outcomes

6 6 36

On throwing two dice at the same time,

Possible outcomes for getting sum as 2

1,1

Possible outcomes for getting sum as 3

2,1 ,

1,2

Possible outcomes for getting sum as 4

3,1 ,

1,3 ,

2,2

Possible outcomes for getting sum as 5

4,1 ,

1,4 ,

2,3 ,

3,2

Possible outcomes for getting sum as 6

5,1 ,

1,5 ,

1,5 ,

2,4 ,

4,2 ,

3,3

Possible outcomes for getting sum as 7

6,1 ,

2,5 ,

5,2 ,

3,4 ,

4,3

Possible outcomes for getting sum as 8

6,2 ,

2,6 ,

3,5 ,

5,3 ,

4,4

Possible outcomes for getting sum as 9

3,6 ,

6,3 ,

4,5 ,

5,4

Possible outcomes for getting sum as 10

4,6 ,

6,4 ,

5,5

Possible outcomes for getting sum as 11

5,6 ,

6,5

5,6 ,

6,5

Possible outcomes for getting sum as 12

6,6

Event:

‘Sum on

2 dice’

2

3

4

5

6

7

8

9

10

11

12

Probabili

ty

1

36

2

36

3

36

4

36

5

36

6

36

5

36

4

36

3

36

2

36

1

36

(ii) Since the sum are not equally likely, probability

of these sums cannot be .

1

11

Question: 23

A game consists of tossing a one rupee coin 3 times

and noting its outcome each time. Hanif wins if all

the tosses give the same result i.e., three heads or

three tails, and loses otherwise. Calculate the

probability that Hanif will lose the game.

Solution

Let head and tail be denoted by and

H

T

respectively.

The possible outcomes are

HHH,TTT,HHT,HTH,THH,TTH,THT,HTT

Number of possible outcomes

8

Number of outcomes having three heads and tails

2

Let be the event of winning of Hanif.

E

Thus,

Number of outcomeshaving threeheadsand tails

P E

Total number of outcomes

2

8

1

4

Probability that Hanif will win

1

4

Probability that Hanif will lose

1 4 1 3

1

4 4 4

Question: 24

A die is thrown twice. What is the probability that

(i) 5 will not come up either time?

(ii) 5 will come up at least once?

Solution

Total outcomes on throwing the dice

6 6 36

(i) Outcomes of having 5 either time can be

written as:

5,1 ,

5,2 ,

5,3 ,

5,4 ,

5,5 ,

5,6 ,

1,5 ,

2,5 ,

3,5 ,

4,5 ,

5,5 ,

6,5

Total number of favourable outcomes

11

Let be the event of having 5 either times.

E

Thus,

Number of favourableoutcomes

P E

Total number of outcomes

11

36

Thus, the probability that 5 will not come up

either time .

11 36 11 25

1

36 36 36

(ii) Total number of outcomes when 5 can come at

least once

11

Number of favourableoutcomes

P E

Total number of outcomes

11

36

Thus, the probability that 5 will come at least

once is .

11

36

Question: 25

Which of the following arguments are correct and

which are not correct? Give reasons for your answer.

(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

Solution

(i) Incorrect

The possible outcomes on tossing two coins

H,H ,

H,T ,

T,H ,

T,T

There are one of each in and .

H,T

T,H

The probability of getting two heads

1

4

The probability of getting two tails

1

4

The probability of getting one of each

1

2

(ii) Correct

The possible outcomes on throwing the dice are

1, 2, 3, 4, 5, and 6.

Odd outcomes are 1, 3, 5 and even outcomes

are 2, 4, 6.

Thus, the probability of getting an odd number

is .

1

2

Exercise 15.2(5)(Optional)

Question: 1

Two customers Shyam and Ekta are visiting a

particular shop in the same week (Tuesday to

Saturday). Each is equally likely to visit the shop on

any day as on another day. What is the probability

that both will visit the shop on

(i) the same day?

(ii) consecutive days?

(iii) different days?

Solution

Total number of days

5

Number of ways for Shyam to go shop

5

Number of ways for Ekta to go shop

5

Total outcomes

5 5 25

(i) Number of ways in which both can reach on

same day

5

Thus, the number of favourable outcomes .

5

Probability of reaching both on same day

Number of favourableoutcomes

Total number of outcomes

5

25

1

5

Thus, the probability of reaching both on same

day is .

1

5

(ii) Outcomes of reaching on consecutive days can

be written as:

Tue,Wed , Wed,Thu , Thu,Fri

, Fri,Sat , Wed,Tue , Thu,Wed ,

Fri,Thu , Sat,Fri ,

The number of ways of reaching on consecutive

days

8

Thus, the number of favourable outcomes .

8

Probability of reaching both on consecutive

days

Number of favourableoutcomes

Total number of outcomes

8

25

Thus, the probability of reaching both on

consecutive days is .

8

25

(iii) Probability of reaching both on same day

1

5

Probability of reaching both on different day

.

1 5 1 4

1

5 5 5

Question: 2

A die is numbered in such a way that its faces show

the numbers 1, 2, 2, 3, 3, 6. It is thrown two times

and the total score in two throws is noted. Complete

the following table which gives a few values of the

total score on the two throws:

Number in first throw

+

1

2

2

3

3

6

1

2

3

3

4

4

7

2

3

4

4

5

5

2

3

Number in Second throw

3

5

9

6

7

8

8

9

9

12

What is the probability that the total score is

(i) even?

(ii) 6?

(iii) at least 6?

Solution

1

2

2

3

3

6

1

2

3

3

4

4

7

2

3

4

4

5

5

8

2

3

4

4

5

5

8

3

4

5

5

6

6

9

3

4

5

5

6

6

9

6

7

8

8

9

9

12

On throwing two dice, possible outcomes

6 6 36

(i) Number of times the sum is even

18

Thus, the number of favorable outcomes .

18

Let be the event that the total score is even.

E

Thus,

Number of favourableoutcomes

P E

Total number of outcomes

18

36

Hence, the probability of having an even

number

18 1

36 2

(ii) Number of times the sum is 6

4

Thus, the number of favourable outcomes .

4

Let be the event that the total score is six.

E

Thus,

Number of favourableoutcomes

P E

Total number of outcomes

4

36

Hence, the probability of having sum as 6

4 1

36 9

(iii) Number of times the sum is at least 6 .

15

Thus, the number of favorable outcomes .

15

Let be the event that the total score is at

E

least six.

Thus,

Number of favourableoutcomes

P E

Total number of outcomes

15

36

Probability of having sum at least 6 .

15 5

36 12

Question: 3

A bag contains 5 red balls and some blue balls. If the

probability of drawing a blue ball is double that of a

red ball, determine the number of blue balls in the

bag.

Solution

Let us consider the number of blue balls

x

Number of red balls

5

Total number of balls

x 5

Probability of having a red ball

Number of red balls

Total number of balls

Probability of having a red ball

5

x 5

Probability of having a blue ball

Number of blueballs

Total number of balls

Probability of having a blue ball

x

x 5

According to the question,

5 x

2

x 5 x 5

2

10 x 5 x 5x

2

x 5x 50 0

2

x 10x 5x 50 0

x x 10 5 x 10 0

x 10 x 5 0

That is,

or

x 10 0

x 5 0

or

x 10

x 5

Number of balls is always positive.

Thus, number of blue balls is .

10

Question: 4

A box contains 12 balls out of which are black. If

x

one ball is drawn at random from the box, what is

the probability that it will be a black ball? If 6 more

black balls are put in the box, the probability of

drawing a black ball is now double of what it was

before. Find

.

x

Solution

Total number of balls

12

Number of black balls

x

Probability of having a black ball

Number of black balls

Total number of balls

Probability of having a black ball

x

12

On putting 6 more black balls,

Total number of balls

12 6 18

Total number of black balls

x 6

Probability of having a black ball

x 6

18

According to the question,

x x 6

2

12 18

3x x 6

2x 6

x 3

Thus, .

x 3

Question: 5

A jar contains 24 marbles, some are green and others

are blue. If a marble is drawn at random from the

jar, the probability that it is green is Find the

2

3

number of blue balls in the jar.

Solution

Total number of marbles in the jar

24

Let us consider the number of green marbles

x

Number of blue marbles

24 x

Numbe

Probability

r of green marb

of having

a green marble

les

Total number of marbles

x

24

According to the question,

x 2

24 3

2 24

x

3

x 2 8

x 16

Thus, the number of green marbles in the jar

16

Total number of blue marbles

24 x

Put the value of in the above expression.

x

Total number of blue marbles

24 16 8

Hence, the number of blue marbles in the jar is 8.