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.