In this post, you will find the NCERT solutions for class 10 maths ex 14.1. These solutions are based on the latest syllabus of NCERT Maths class 10.
NCERT Solutions for Class 10 Maths Ex 14.1
1. Complete the following
statements:
(i)
Probability of an event E + Probability of event ‘not E’ = 1.
(ii) The
probability of an event that cannot happen is 0. Such an
event is called impossible event.
(iii) The
probability of an event that is certain to happen is 1.
Such an event is called 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.
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) In the experiment, “A driver attempts to start a car. The
car starts or does not start”, it is not sure that each outcome is as likely to
occur as the other. Thus, the experiment has no equally likely outcomes.
(ii) In
the experiment, “A player attempts to shoot a basketball. She/he shoots or
misses the shot”, it is not sure that each outcome is as likely to occur as the
other. Thus, the experiment has no equally likely outcomes.
(iii) In
the experiment “A trial is made to answer a true-false question. The answer is
right or wrong”, it is clear that the result can lead in one of the two
possible answers – either right or wrong. We can clearly justify that each
outcome, right or wrong, is likely to occur as the other. Thus, the experiment
has equally likely outcomes.
(iv) In
the experiment, “A baby is born. It is a boy or a girl”, it is clear that the
outcome can lead in one of the two possible outcomes – either a boy or a girl.
We can clearly justify that each outcome, a boy or a girl, is likely to occur
as the other. Thus, the experiment has equally likely outcomes.
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:
The tossing of a
coin is considered to be a fair way of deciding which team should get the ball
at the beginning of a football game because it is clear that the tossing of the
coin only land in one of two possible sides – either head up or tail up. It can
be justified that each outcome, head or tail, is as likely to occur as the
other, i.e., the outcomes head and tail are equally likely. So, the result of
the tossing a coin is completely unpredictable.
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) Since the
probability of an event E is a number P(E) such that
0 ≤ P(E) ≤ 1
It means that the
probability of an event is greater or equal to 0 but less than or equal to 1.
Here, –1.5 is less than 0
because of negative sign.
Therefore, –1.5
cannot be the
probability of an event.
5. If P(E) = 0.05, what is the probability of
‘not E’?
Solution: We know that, P(E) + P(not E) = 1
or P(not
E) = 1 – P(E)
= 1 – 0.05
= 0.95
Therefore,
the probability of ‘not E’ is 0.95.
6. A bag contains lemon flavoured candies
only. Malini takes out one candy without looking into the bag. What is the
probability that she takes out
(i) an orange flavoured candy?
(ii) a lemon flavoured candy?
Solution:
(i) Consider
the event E related to the experiment of taking out of an orange flavoured
candy from a bag containing only lemon flavoured candies. Since no outcome
gives an orange flavoured candy, therefore, it is an impossible event. So, its
probability is 0. Hence, P(E) = 0.
(ii) Consider the event E of taking a
lemon flavoured candy out from a bag containing only lemon flavoured candies.
This event is a sure or a certain event. So, its probability is 1. Hence, P(E)
= 1.
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: Let E be the event of having the same
birthday, then ‘not E’ is the event of not having the same birthday.
It is given that, P(not E) = 0.992
But
P(E) + P(not E) = 1
P(E) = 1 – P(not E)
= 1 – 0.992
= 0.008
Hence,
the probability that the 2 students have the same birthday is 0.008.
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:
There are a total
of 3 + 5 = 8 balls in a bag. Out of these 8 balls, one can be drawn in 8 ways.
Total
number of possible outcomes = 8
(i) Since the bag contains 3 red balls,
therefore, one red ball can be drawn in 3 ways.
Number of favourable outcomes = 3
Hence, P(getting a red ball) = 3/8
(ii) Since the bag contains 5 black balls
along with 3 red balls, therefore, 5 balls are not red balls. One black (not
red) ball can be drawn in 5 ways.
Number
of favourable outcomes = 5
Hence,
P(getting not a red ball) = 5/8
Note: We can simply solve
this question as follows:
If P(E) denotes the probability of getting a red ball, then
the probability of getting not a red ball, i.e., P(not E) can be calculated by
the formula P(not E) = 1 – P(E).
Thus, P(getting not a red ball) = 1 – P(getting
a red ball)
P(getting not a red ball) = 1 – 3/8 = 5/8
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 in the box = 5
+ 8 + 4 = 17
Total
number of possible outcomes = 17
(i) There are 5 red marbles in the box.
Number
of favourable outcomes = 5
P(getting
a red marble) = 5/17
(ii) There are 8 white marbles in the box.
Number
of favourable outcomes = 8
P(getting
a white marble) = 8/17
(iii) There are 5 + 8 = 13 marbles in the box, which
are not green.
Number
of favourable outcomes = 13
P(getting
not a green marble) = 13/17
10. A piggy bank contains hundred 50 p 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 50 p coin?
(ii) will not be a Rs.5 coin?
Solution:
Total number of
coins in the piggy bank = 100 + 50 + 20 + 10 = 180
Total
number of possible outcomes = 180
(i) There are one hundred 50 p coins in the piggy
bank.
Number
of favourable outcomes = 100
P(falling out a 50 p coin) = 100/180 = 5/9
(ii) There are 100 + 50 + 20 = 170 coins, which are
not Rs. 5 coins.
Number
of favourable outcomes = 170
So, P(falling
out a coin which is not a Rs. 5 coin) = 170/180 = 17/18
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 figure). What is the probability that the fish
taken out is a male fish?
Solution: Total number of fish in the tank = 5 +
8 = 13
Total
number of possible outcomes = 13
There
are 5 male fish in the tank.
Number
of favourable outcomes = 5
Hence,
P(taking out a male fish) = 5/13
12. A game of chance consists of spinning an
arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8
(see figure), and these are equally 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?
Solution:
Out of 8 numbers,
an arrow can point any of the numbers in 8 ways.
Total
number of possible outcomes = 8
(i) Since there
are only one 8.
Number of favourable outcomes = 1
Hence,
P(arrow points at 8) = 1/8
(ii) Since there are 4
odd numbers, i.e., 1, 3, 5, 7.
Number of favourable outcomes = 4
Hence,
P(arrow points at an odd number) = 4/8 =
1/2
(iii) Since the numbers
greater than 2 are 3, 4, 5, 6, 7, 8.
Number of favourable outcomes = 6
Hence,
P(arrow points at a number > 2) = 6/8 =
3/4
(iv) Since all the 8
numbers are less than 9.
Number of favourable outcomes = 8
Hence,
P(arrow points at a number < 9) = 8/8 =
1
13. A dice 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: Total number of possible outcomes when
throwing a dice = 6
(i) On a dice, the prime numbers are 2, 3
and 5.
Therefore,
the number of favourable outcomes = 3
Hence,
P(getting a prime number) = 3/6 = 1/2
(ii) On a dice, the number lying between 2
and 6 are 3, 4, 5.
Therefore,
the number of favourable outcomes = 3
Hence,
P(getting a number lying between 2 and 6) = 3/6 =1/2
(iii) On a dice, the odd numbers are 1, 3 and 5.
Therefore,
the number of favourable outcomes = 3
Hence,
P(getting an odd number) = 3/6 = 1/2
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: Total number of possible outcomes = 52
(i) There are two suits of red cards,
i.e., diamond and heart. Each suit contains one king. So, there are 2 kings of
red colour.
Number
of favourable outcomes = 2
Hence,
P(a king of red colour) = 2/52 = 1/26
(ii) Since there are 3 face cards in each suit,
therefore, a total of 12 face cards are there in all four suits.
Number
of favourable outcomes = 12
Hence,
P(a face card) = 12/52 = 3/13
(iii) There are two suits of red cards,
i.e., diamond and heart. Each suit contains 3 face cards.
Number
of favourable outcomes = 2 × 3 = 6
Hence,
P(a red face card) = 6/52 = 3/26
(iv) There is only one jack of heart.
Number
of favourable outcomes = 1
Hence,
P(the jack of hearts) = 1/52
(v) There are 13 cards of spade.
Number
of favourable outcomes = 13
Hence,
P(a spade) = 13/52 = 1/4
(vi) There is only one queen of diamonds.
Number
of favourable outcomes = 1
Hence,
P(the queen of diamonds) = 1/52
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: Total number of possible outcomes = 5
(i) There is only one queen.
Number
of favourable outcomes = 1
Hence,
P(the queen) = 1/5
(ii) If the queen is put aside, total number of possible
outcomes = 4
(a)
Number of favourable outcomes = 1
Hence,
P(an ace) = 1/4
(b)
Since the queen is put aside, there is no card as queen.
Number
of favourable outcomes = 0
Hence,
P(the queen) = 0/4 = 0
16. 12 defective pens are accidently 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: Total number of possible outcomes =
132 + 12 = 144
Number
of favourable outcomes = 132
Hence,
P(getting a good pen) = 132/144 = 11/12
17. (i) A lot of 20 bulbs contains 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: (i) Total number of possible
outcomes = 20
Number
of favourable outcomes = 4
Hence,
P(getting a defective bulb) = 4/20 = 1/5
(ii) If the bulb drawn in (i) is not
defective, then in the lot, number of defective bulb is 4 and number of not
defective bulb is 15.
Now, total number of possible outcomes = 20 –
1 = 19
Number
of favourable outcomes = 19 – 4 = 15
Hence,
P(getting a non-defective bulb) = 15/19
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: Total number of possible outcomes = 90
(i) Number of two-digit numbers from 1 to 90 are
90 – 9 = 81
Number
of favourable outcomes = 81
Hence,
P(getting a disc bearing a two-digit number) = 81/90 = 9/10
(ii) From 1 to 90, the perfect square numbers are
1, 4, 9, 16, 25, 36, 49, 64 and 81.
Number
of favourable outcomes = 9
Hence,
P(getting a perfect square number) = 9/90 = 1/10
(iii) From 1 to 90, the numbers divisible by 5 are
5, 10, 15, 20, ……., 90
There are total 18 numbers which are divisible
by 5.
Number
of favourable outcomes = 18
Hence,
P(getting a number divisible by 5) = 18/90 = 1/5
19. A child has a die whose six faces show the
letters as given below:
A B C
D E A
The die is thrown once. What is the
probability of getting
(i) A?
(ii) D?
Solution: Total number of possible outcomes = 6
(i) Since the letter A
appears on the two faces.
Number of favourable outcomes = 2
Hence,
P(getting a letter A) = 2/6 = 1/3
(ii) Since the letter D
appears on only one face.
Number of favourable outcomes = 1
Hence,
P(getting a letter D) = 1/6
20. Suppose you drop a die at random on the
rectangular region shown in the following figure. What is the probability that
it will land inside the circle with diameter 1 m?
Solution:
Area of the rectangle
= 3 × 2 = 6 m2
And area of the circle = πr2 = π(1/2)2 = π/4 m2
Total possible area to land the die = 6 m2
Total favourable area to land the die = π/4 m2
Hence,
P(die to land inside the circle) = π/4/6 = π/24
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 possible outcomes = 144
(i) Number of non-defective pens = 144 – 20 = 124
Number
of favourable outcomes = 124
Hence,
P(she will buy) = P(a non-defective pen) = 124/144 = 31/36
(ii) Number of favourable outcomes = 20
Hence,
P(she will not buy) = P(a defective pen) = 20/144 = 5/36
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.
Solution: Total possible outcomes of throwing two dice
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)
Total
number of possible outcomes = 36
(i) Favourable outcomes of getting the sum as 3,
i.e., (1, 2) and (2, 1) = 2
Hence,
P(getting the sum as 3) = 2/36 = 1/18
Favourable
outcomes of getting the sum as 4, i.e., (1, 3), (3, 1) and (2, 2) = 3
Hence,
P(getting the sum as 4) = 3/36 = 1/12
Favourable
outcomes of getting the sum as 5 = 4
Hence,
P(getting the sum as 5) = 4/36 = 1/9
Favourable
outcomes of getting the sum as 6 = 5
Hence,
P(getting the sum as 6) = 5/36
Favourable
outcomes of getting the sum as 7 = 6
Hence,
P(getting the sum as 7) = 6/36 = 1/6
Favourable
outcomes of getting the sum as 9 = 4
Hence,
P(getting the sum as 9) = 4/36 = 1/9
Favourable
outcomes of getting the sum as 10 = 3
Hence,
P(getting the sum as 10) = 3/36 = 1/12
Favourable
outcomes of getting the sum as 11 = 2
Hence,
P(getting the sum as 11) = 2/36 = 1/18
|
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) I do not agree with the argument given
by the student. Justification has already been given in part (i).
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: The outcomes when a coin is tossed 3 times:
HHH,
HHT, HTH, THH, TTH, HTT, THT, TTT
Total
number of possible outcomes = 8
Number
of favourable outcomes = 6
Hence,
P(Hanif will lose the game) = 6/8 = 3/4
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:
(i) The
outcomes when a dice is thrown is twice:
(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)
Total
number of possible outcomes = 36
Now,
consider the event E in which 5 will come up either time.
Then, E = (5, 1) (5, 2) (5, 3) (5,
4) (5, 5) (5, 6) (1, 5) (2, 5) (3, 5) (4, 5) (6, 5)
Total
number of outcomes favourable to E = 11
P(5
will come up either time) = P(E) = 11/36
P(5
will not come up either time) = P(not E) = 1 – 11/36 = 25/36
(ii) Total number of possible outcomes = 36
Now,
consider the event E in which 5 will come up at least once.
Then, E = (5, 1) (5, 2) (5, 3) (5,
4) (5, 5) (5, 6) (1, 5) (2, 5) (3, 5) (4, 5) (6, 5)
Total
number of outcomes favourable to E = 11
P(5
will come up at least once) = P(E) = 11/36
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: We
can classify the outcomes like this but they are not then, ‘equally likely’.
Reason is that ‘one of each’ can result in two ways – from a head on first coin
and tail on the second coin or from a tail on the first coin and head on the
second coin. This makes it twice as likely as two heads (or two tails).
(ii) Correct: The
two outcomes considered in the question are equally likely.