Probability - Question Bank for Class/Grade 10 maths.

Probability
1) A bag contains a green ball, a white ball and a black ball all balls being of the
same shape and size. Rohan takes a ball from the bag without looking into it.
What is the probability that he takes out a black ball??
Solution:
Since Rohan takes the ball out without looking into it. So it is equally likely
that he takes out any one of them.
Since there is only one black ball,
Let B be the event that we get a black ball
The number of favourable outcomes = 1
The number of possible outcomes = 3
P(B)=
The number of favourable outcomes
The number of possible outcomes
1
P(B) =
3
2) Suppose, we throw a die once,
(i) What is the probability that we get a number greater than 3
(ii) What is the probability of getting a number less than or equal to 3?
Solution:
Let E be an event that we get a number greater than 3.
The possibilities according to given condition are 4, 5, 6.
The total possibilities are 1, 2, 3,4,5,6.
P(E) =
=
3
6
1
P(E) =
2
Let D be the event that we get a number less than or equal to 3.
The given possibilities are 1, 2, 3.
The total possibilities are 1, 2, 3, 4, 5, 6.

Probability
P(E) =
=
3
6
1
P(E) =
2
3) A single card is drawn from a well shuffled pack of 52 cards.
Calculate the probability that the card is
(i) a King
(ii) not a king
Solution:
(i) Let E be the event that the card is a King.
There are 4 kings in a deck
P (E) =
=
4
52
1
P(E) =
13
(ii) Let D be the event that the card is not a king
Hence D = E
Therefore,
P( E ) = 1- P(E)
= 1-
1
13
P( E ) =
12
13
12
P(D) =
13

Probability
4) Two players, Rahul and Rohit are playing a Badminton match.
The probability of Rahul winning the match is 0.37.
What is the probability of Rohit winning the match?
Solution:
Let E denote the event that Rahul wins the match and
D denote the event that Rohit wins the match.
Given that,
Probability that Rahul wins the match = P(E) = 0.37
Therefore
Probability that Rohit wins the match = P(D) = 1- P(E)
Probability that Rohit wins the match = P(D) = 1- 0.37 = 0.63
5) There are 60 students in class XII of a school of whom 35 are boys.
The class teacher has to select a class monitor. She tell the students to write
their names on an identical piece of paper. Then she put all the chits in a
bowl and stirs it thoroughly and draws a card at random from the bowl.
What is the probability that the name written on the chit is that of
(i) a girl (ii) a boy?
Solution:
(i)
Let G be the event that the name drawn is that of a girl
P(G) =
P(G) =
25
60
=
5
12
5
P(G) =
12
(ii)
Let B be the event that the name drawn is that of a boy
P(B) = 1 – P(G)

Probability
P(B) = 1 -
5
12
7
P(B) =
12
5) A bas contains only yellow marbles. Avinash takes one marble without looking
into the bag. What is the probability that he takes out
(i) a pink marble (ii) a black marble
Solution:
(i)
Let P be the event that a pink marble is taken out
Since the bag contains no pink marbles
P (P) =
P(P) =0
(ii)
Let Y be the event that a black marble is taken out
Since the bag only contains yellow marbles & no black marble.
( )
6) A box contains 8 green marbles, 3 white marbles and 9 orange marbles.
One marble is taken out of the box at random. What is the probability that
the marble taken out will be (i) green (ii) not white (iii) orange.
Solution:
(i)
Let G be the event that a green marble is taken out.
P (G) =
P (G) =
8
20

Probability
2
P(G) =
5
(ii)
Let E be the event that a marble is taken out which is not white
i.e it can be green or orange
The number of favourable outcomes for green & orange marbles = 17
P(E) =
P(E) =
17
20
17
P(E) =
20
(iii)
Let O be the event that an orange marble is taken out.
P (O) =
9
P(O) =
20
7) A bag contains 16 candies out of which x are orange flavoured
(i) If one candy is drawn at random, what is the probability that it will be
Orange flavoured.
(ii) If 8 more orange flavoured candies are added to the bag, the probability of
taking out an orange flavoured candy is doubled than that in (i). Find x.
Solution:
(i)
Let O be the event that the candy taken out is orange flavoured
The number of favourable outcomes = x

Probability
P(O) =
P(O) =
16
x
(ii)
The number of possible outcomes = 16+8 = 24
Let E be the event that the candy taken out is orange flavoured
The number of favourable outcomes = x+8
P(E) =
P(E) =
8
24
x 
Given that ,
P(E) = 2 x P(O)
8
24
x 
= 2
16
x
16(x+8) = 48x
128 = 48x-16x
32x= 128
x= 4
8) There are 600 toys in a bag out of which 144 are defective. One toy is
picked at random. What is the probability that it is non-defective?
Solution:
The number of possible outcomes =600
Let E be the event that the toy is non-defective
Total number of non-defective toys = 600-144 = 456
Therefore, total number of favourable outcome = 456
P(E) =
P(E) =
456
600
19
P(E) =
25

Probability
9) A bag contains 8 blue balls and some white balls. If the probability of drawing
a white ball is double that of a blue ball. Find the number of white balls in the bag.
Solution:
Let the total number of possible outcomes be n(S)
Let the number of white balls be x
Let B be the event that a blue ball is drawn and W be the event that a white
ball is drawn.
P(W) =
=
( )
x
n S
P(B) =
The number of favourable outcomes 8
The number of possible outcomes ( )n S

Given that,
P(W) = 2 P(B)
Substituting the values we get,
( )
x
n S
= 2
8
( )n S
x = 2 x8
x=16
Hence, the number of white balls = x = 16.
10) Find the probability of getting 52 Sundays in a leap year.
Solution:
Let S be the event that we get 52 Sundays
Number of days in a leap year = 365 + 1 = 366
Hence a leap year will have 52 complete weeks and two days extra
(52 x 7 = 364 + 2 =366)
These two extra days can be (Saturday – Sunday, Sunday – Monday, Monday – Tuesday,
Tuesday-Wednesday, Wednesday – Thursday, Thursday- Friday, Friday- Saturday)
The Total number of possible outcomes = 7
Out of these 7, 5 don’t have a Sunday in them (Monday – Tuesday,
Tuesday-Wednesday, Wednesday – Thursday, Thursday- Friday, Friday- Saturday)

Probability
P(S) =
5
P(S) =
7
11) Sukhdeep tosses two different coins simultaneously. What is the probability
that she gets at least one tail?
Solution:
Let T be the event that we get atleast one tail.
When two coins are tossed simultaneously the possible outcomes are (H, H)
(H, T), (T, H) & (T, T)
Thus The total number of possible outcomes = 4
Out of these 4, 3 have atleast one tail (H, T), (T, H) & (T,T)
P(T) =
The total number of possible outcomes
3
P(T) =
4
12) A warehouse contains 10,000 mobiles. Out of which 7800 are good,
1600 have a small defect and the rest have a major defect. Niki a trader will accept
only those mobiles which are good but ritu, another trader, will only reject
those mobiles which have major defects One mobile is drawn at random.
What is the probability that it is acceptable to
(i) Niki (ii) Ritu
Solution:
The total number of possible outcomes = 10,000
(i)
Let N be the event that the mobile is acceptable to niki

Probability
P(N) =
P(N) =
7,800
10,000
P(N) = 0.78
(ii)
Let R be the event that it is acceptable to ritu.
The number of favourable outcomes = 7800+1600 = 9400
P(R) =
P(R) =
9,400
10,000
P(R) = 0.94
13) Two dice, one black and one green are throw simultaneously. What is
the probability that the sum of the two numbers appearing on the top of
the dice is (i) 4 (ii) 14
Solution:
Since two dice are thrown simultaneously
The different possible 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)
(i)

Probability
Let E be the event that the sum of the numbers appearing on the
top of the dice is 4
The different outcomes such that the sum of the two numbers = 4 are
{ (1,3) (3,1) (2,2)}
P(E) =
=
3
36
1
P(E) =
12
(ii)
Let D be the event that the sum of the numbers appearing on the
top of the dice is 14
There are no outcomes in which the sum of the two numbers appearing
on the top of the two dice = 14
P(D) =
=
0
36
P(D) = 0
14) A wallet contains hundred one rupee coins, 75 two rupee coins
15 five rupee coins and ten ten rupee coins. One coin drops off from the wallet
What is the probability that the coin is (i) a two rupee coin (ii) not a ten rupee coin.
Solution:
The number of possible outcomes = 100 + 75 + 15 + 10 = 200
(i)

Probability
Let E be the event that the coin is a two rupee coin
P(E) =
=
75
200
3
P(E) =
8
(ii)
Let D be the event that it is not a ten rupee coins
Number of ten rupee coins = 10
The number of favourable outcomes = 200 – 10 = 190
P(D) =
=
190
200
19
P(D) =
20
15) A bag contains 100 cards. On each card the numbers 1- 100 are written.
Jack picks out a card from the bag. What is the probability that the number
On the card
(i) Contains a “0” (ii) is divisible by 7
Solution:
The total number of possible outcomes = 100
(i)
Let Z be the event that the number on the card contains a “0”
The favourable outcomes are { 10,20,30,40,50,60,70,80,90,100}

Probability
P(Z) =
=
10
100
1
P(Z) =
10
(ii)
Let S be the event that the number on the card is divisible by 7
The favourable outcomes are { 7,14,21,28,35,42,49,56,63,70,77,84,91,98}
P(S) =
=
14
100
7
P(S) =
50
16) Five cards - the ten, Jack, Queen, King and Ace of Spades are well shuffled
With their face downwards one card is then picked up at random.
(I) What is the probability that it is Jack
(II) If Jack is drawn and put aside, What is the probability that the second card drawn
a king
Solution:
The total number of possible outcomes = 5
(I)
Let J be the event that Jack is drawn
P(J) =

1
P(S) =
5
(II)
Since Jack is drawn and put aside, only 4 cards are remaining.
Therefore, The number of possible outcomes = 4
Let K be the probability that a king is drawn
P(K) =
1
P(K) =
4
17) A die is thrown once. Find the probability of getting
(i) a composite number (ii) a number lying between 1 and 6
(iii) an even number
Solution:
(i)
Let C be the event that the number is a composite number
The favourable outcomes are { 4, 6}
P(C) =
=
2
6
1
P(C) =
3

Probability
(ii)
Let N be the event that the number is lying between 1 and 6
The favourable outcomes are {2, 3, 4, 5}
P (N) =
=
4
6
P (N) =
(iii) Let E be the event that the number is an even number
The favourable outcomes are {2, 4,6}
P (E) =
=
P (E) =
18) A die is thrown twice. What is the probability that
(i) 6 will not come either time (ii) 4 will come up atleast once
Solution:
The different possible 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)

Probability
(i)
Let S be the event that 6 will not come either time
The favourable outcomes are { (1,1) (1,2) (1,3) (1,4) (1,5) (2,1) (2,2)
(2,3) (2,4) (2,5)(3,1) (3,2) (3,3) (3,4) (3,5) (4,1) (4,2) (4,3) (4,4) (4,5)
(5,1) (5,2) (5,3) (5,4) (5,5)}
P(S) =
25
P(S) =
36
(ii)
Let F be the event that 4 will come up at least once
The favourable outcomes are { (1,4) (2,4) (3,4) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,4) (6,4)}
P(F) =
=
11
36
11
P(F) =
36
19) The probability of a team winning the match is
15
p
. If the match cannot end in a tie,
and the probability of the team losing the match is
2
5
. What is the value of p?
Solution:
Let W be the event that the team won the match and L be the event that the team
Lost the match.
Since the match cannot end in a tie.

P(W) + P(L) = 1
15
p
+
2
5
= 1
15
p
= 1 -
2
5
=
3
5
p = 15 x
3
5
9p 
20) A number is chosen at random from the numbers 2,5,5,6,6,6,8,8,8,10,10,12,12,24,28
What is the probability that the selected number is their average?
Solution:
The average of the numbers =
2 5 5 6 6 6 8 8 8 10 10 12 12 24 28
15
              
 
 
The average of the numbers = 10
Let E be the event that “10” is selected from the numbers.
P(E) =
=
2
15
2
P(E) =
15

Probability - Question Bank for Class/Grade 10 maths.

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Probability - Question Bank for Class/Grade 10 maths.