Ncert class-10-mathematics-chapter-15

PROBABILITY 295
15
The theory of probabilities and the theory of errors now constitute
a formidable body of great mathematical interest and of great
practical importance.
– R.S. Woodward
15.1 Introduction
In Class IX, you have studied about experimental (or empirical) probabilities of events
which were based on the results of actual experiments. We discussed an experiment
of tossing a coin 1000 times in which the frequencies of the outcomes were as follows:
Head : 455 Tail : 545
Based on this experiment, the empirical probability of a head is
455
1000
, i.e., 0.455 and
that of getting a tail is 0.545. (Also see Example 1, Chapter 15 of Class IX Mathematics
Textbook.) Note that these probabilities are based on the results of an actual experiment
of tossing a coin 1000 times. For this reason, they are called experimental or empirical
probabilities. In fact, experimental probabilities are based on the results of actual
experiments and adequate recordings of the happening of the events. Moreover,
these probabilities are only ‘estimates’. If we perform the same experiment for another
1000 times, we may get different data giving different probability estimates.
In Class IX, you tossed a coin many times and noted the number of times it turned up
heads (or tails) (refer to Activities 1 and 2 of Chapter 15). You also noted that as the
number of tosses of the coin increased, the experimental probability of getting a head
(or tail) came closer and closer to the number
1
2
⋅ Not only you, but many other
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296 MATHEMATICS
persons from different parts of the world have done this kind of experiment and recorded
the number of heads that turned up.
For example, the eighteenth century French naturalist Comte de Buffon tossed a
coin 4040 times and got 2048 heads. The experimental probabilility of getting a head,
in this case, was 2048
4040
, i.e., 0.507. J.E. Kerrich, from Britain, recorded 5067 heads in
10000 tosses of a coin. The experimental probability of getting a head, in this case,
was
5067
0.5067
10000
= . Statistician Karl Pearson spent some more time, making 24000
tosses of a coin. He got 12012 heads, and thus, the experimental probability of a head
obtained by him was 0.5005.
Now, suppose we ask, ‘What will the experimental probability of a head be if the
experiment is carried on upto, say, one million times? Or 10 million times?And so on?’
You would intuitively feel that as the number of tosses increases, the experimental
probability of a head (or a tail) seems to be settling down around the number 0.5 , i.e.,
1
2
, which is what we call the theoretical probability of getting a head (or getting a
tail), as you will see in the next section. In this chapter, we provide an introduction to
the theoretical (also called classical) probability of an event, and discuss simple problems
based on this concept.
15.2 Probability — A Theoretical Approach
Let us consider the following situation :
Suppose a coin is tossed at random.
When we speak of a coin, we assume it to be ‘fair’, that is, it is symmetrical so
that there is no reason for it to come down more often on one side than the other.
We call this property of the coin as being ‘unbiased’. By the phrase ‘random toss’,
we mean that the coin is allowed to fall freely without any bias or interference.
We know, in advance, that the coin can only land in one of two possible ways —
either head up or tail up (we dismiss the possibility of its ‘landing’ on its edge, which
may be possible, for example, if it falls on sand). We can reasonably assume that each
outcome, head or tail, is as likely to occur as the other. We refer to this by saying that
the outcomes head and tail, are equally likely.
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PROBABILITY 297
For another example of equally likely outcomes, suppose we throw a die
once. For us, a die will always mean a fair die. What are the possible outcomes?
They are 1, 2, 3, 4, 5, 6. Each number has the same possibility of showing up. So
the equally likely outcomes of throwing a die are 1, 2, 3, 4, 5 and 6.
Are the outcomes of every experiment equally likely? Let us see.
Suppose that a bag contains 4 red balls and 1 blue ball, and you draw a ball
without looking into the bag. What are the outcomes? Are the outcomes — a red ball
and a blue ball equally likely? Since there are 4 red balls and only one blue ball, you
would agree that you are more likely to get a red ball than a blue ball. So, the outcomes
(a red ball or a blue ball) are not equally likely. However, the outcome of drawing a
ball of any colour from the bag is equally likely. So, all experiments do not necessarily
have equally likely outcomes.
However, in this chapter, from now on, we will assume that all the experiments
have equally likely outcomes.
In Class IX, we defined the experimental or empirical probability P(E) of an
event E as
P(E) =
Number of trials in which the event happened
Total number of trials
The empirical interpretation of probability can be applied to every event associated
with an experiment which can be repeated a large number of times. The requirement
of repeating an experiment has some limitations, as it may be very expensive or
unfeasible in many situations. Of course, it worked well in coin tossing or die throwing
experiments. But how about repeating the experiment of launching a satellite in order
to compute the empirical probability of its failure during launching, or the repetition of
the phenomenon of an earthquake to compute the empirical probability of a multi-
storeyed building getting destroyed in an earthquake?
In experiments where we are prepared to make certain assumptions, the repetition
of an experiment can be avoided, as the assumptions help in directly calculating the
exact (theoretical) probability. The assumption of equally likely outcomes (which is
valid in many experiments, as in the two examples above, of a coin and of a die) is one
such assumption that leads us to the following definition of probability of an event.
The theoretical probability (also called classical probability) of an event E,
written as P(E), is defined as
P(E) =
Number of outcomes favourable to E
Number of all possible outcomes of the experiment
,
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298 MATHEMATICS
where we assume that the outcomes of the experiment are equally likely.
We will briefly refer to theoretical probability as probability.
This definition of probability was given by Pierre Simon Laplace in 1795.
Probability theory had its origin in the 16th century when
an Italian physician and mathematician J.Cardan wrote the
first book on the subject, The Book on Games of Chance.
Since its inception, the study of probability has attracted
the attention of great mathematicians. James Bernoulli
(1654 – 1705), A. de Moivre (1667 – 1754), and
Pierre Simon Laplace are among those who made significant
contributions to this field. Laplace’s Theorie Analytique
des Probabilités, 1812, is considered to be the greatest
contribution by a single person to the theory of probability.
In recent years, probability has been used extensively in
many areas such as biology, economics, genetics, physics,
sociology etc.
Let us find the probability for some of the events associated with experiments
where the equally likely assumption holds.
Example 1 : Find the probability of getting a head when a coin is tossed once. Also
find the probability of getting a tail.
Solution : In the experiment of tossing a coin once, the number of possible outcomes
is two — Head (H) and Tail (T). Let E be the event ‘getting a head’. The number of
outcomes favourable to E, (i.e., of getting a head) is 1. Therefore,
P(E) = P (head) =
Number of all possible outcomes
=
1
2
Similarly, if F is the event ‘getting a tail’, then
P(F) = P(tail) =
1
2
(Why ?)
Example 2 : A bag contains a red ball, a blue ball and a yellow ball, all the balls being
of the same size. Kritika takes out a ball from the bag without looking into it. What is
the probability that she takes out the
(i) yellow ball? (ii) red ball? (iii) blue ball?
Pierre Simon Laplace
(1749 – 1827)
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PROBABILITY 299
Solution : Kritika takes out a ball from the bag without looking into it. So, it is equally
likely that she takes out any one of them.
Let Y be the event ‘the ball taken out is yellow’, B be the event ‘the ball taken
out is blue’, and R be the event ‘the ball taken out is red’.
Now, the number of possible outcomes = 3.
(i) The number of outcomes favourable to the event Y = 1.
So, P(Y) =
1
3
Similarly, (ii) P(R) =
1
3
and (iii) P(B) =
1
3
⋅
Remarks :
1. An event having only one outcome of the experiment is called an elementary
event. In Example 1, both the events E and F are elementary events. Similarly, in
Example 2, all the three events, Y, B and R are elementary events.
2. In Example 1, we note that : P(E) + P(F) = 1
In Example 2, we note that : P(Y) + P(R) + P(B) = 1
Observe that the sum of the probabilities of all the elementary events of
an experiment is 1. This is true in general also.
Example 3 : Suppose we throw a die once. (i) What is the probability of getting a
number greater than 4 ? (ii) What is the probability of getting a number less than or
equal to 4 ?
Solution : (i) Here, let E be the event ‘getting a number greater than 4’. The number
of possible outcomes is six : 1, 2, 3, 4, 5 and 6, and the outcomes favourable to E are 5
and 6. Therefore, the number of outcomes favourable to E is 2. So,
P(E) = P(number greater than 4) =
2
6
=
1
3
(ii) Let F be the event ‘getting a number less than or equal to 4’.
Number of possible outcomes = 6
Outcomes favourable to the event F are 1, 2, 3, 4.
So, the number of outcomes favourable to F is 4.
Therefore, P(F) =
4
6
=
2
3
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Are the events E and F in the example above elementary events? No, they are
not because the event E has 2 outcomes and the event F has 4 outcomes.
Remarks : From Example 1, we note that
P(E) + P(F) =
1 1
1
2 2
+ = (1)
where E is the event ‘getting a head’ and F is the event ‘getting a tail’.
From (i) and (ii) of Example 3, we also get
P(E) + P(F) =
1 2
1
3 3
+ = (2)
where E is the event ‘getting a number >4’ and F is the event ‘getting a number ≤ 4’.
Note that getting a number not greater than 4 is same as getting a number less
than or equal to 4, and vice versa.
In (1) and (2) above, is F not the same as ‘not E’? Yes, it is. We denote the event
‘not E’ by E .
So, P(E) + P(not E) = 1
i.e., P(E) + P( E ) = 1, which gives us P( E ) = 1 – P(E).
In general, it is true that for an event E,
P( E ) = 1 – P(E)
The event E , representing ‘not E’, is called the complement of the event E.
We also say that E and E are complementary events.
Before proceeding further, let us try to find the answers to the following questions:
(i) What is the probability of getting a number 8 in a single throw of a die?
(ii) What is the probability of getting a number less than 7 in a single throw of a die?
Let us answer (i) :
We know that there are only six possible outcomes in a single throw of a die. These
outcomes are 1, 2, 3, 4, 5 and 6. Since no face of the die is marked 8, so there is no
outcome favourable to 8, i.e., the number of such outcomes is zero. In other words,
getting 8 in a single throw of a die, is impossible.
So, P(getting 8) =
0
6
= 0
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PROBABILITY 301
That is, the probability of an event which is impossible to occur is 0. Such an
event is called an impossible event.
Let us answer (ii) :
Since every face of a die is marked with a number less than 7, it is sure that we
will always get a number less than 7 when it is thrown once. So, the number of
favourable outcomes is the same as the number of all possible outcomes, which is 6.
Therefore, P(E) = P(getting a number less than 7) =
6
6
= 1
So, the probability of an event which is sure (or certain) to occur is 1. Such an event
is called a sure event or a certain event.
Note : From the definition of the probability P(E), we see that the numerator (number
of outcomes favourable to the event E) is always less than or equal to the denominator
(the number of all possible outcomes). Therefore,
0 ≤≤≤≤≤ P(E) ≤≤≤≤≤ 1
Now, let us take an example related to playing cards. Have you seen a deck of
playing cards? It consists of 52 cards which are divided into 4 suits of 13 cards each—
spades (m), hearts (n), diamonds (o) and clubs (p). Clubs and spades are of black
colour, while hearts and diamonds are of red colour. The cards in each suit are ace,
king, queen, jack, 10, 9, 8, 7, 6, 5, 4, 3 and 2. Kings, queens and jacks are called face
cards.
Example 4 : One card is drawn from a well-shuffled deck of 52 cards. Calculate the
probability that the card will
(i) be an ace,
(ii) not be an ace.
Solution : Well-shuffling ensures equally likely outcomes.
(i) There are 4 aces in a deck. Let E be the event ‘the card is an ace’.
The number of outcomes favourable to E = 4
The number of possible outcomes = 52 (Why ?)
Therefore, P(E) =
4 1
52 13
=
(ii) Let F be the event ‘card drawn is not an ace’.
The number of outcomes favourable to the event F = 52 – 4 = 48 (Why?)
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302 MATHEMATICS
The number of possible outcomes = 52
Therefore, P(F) =
48 12
52 13
=
Remark : Note that F is nothing but E . Therefore, we can also calculate P(F) as
follows: P(F) = P( E ) = 1 – P(E) =
1 12
1
13 13
− = ⋅
Example 5 : Two players, Sangeeta and Reshma, play a tennis match. It is known
that the probability of Sangeeta winning the match is 0.62. What is the probability of
Reshma winning the match?
Solution : Let S and R denote the events that Sangeeta wins the match and Reshma
wins the match, respectively.
The probability of Sangeeta’s winning = P(S) = 0.62 (given)
The probability of Reshma’s winning = P(R) = 1 – P(S)
[As the events R and S are complementary]
= 1 – 0.62 = 0.38
Example 6 : Savita and Hamida are friends. What is the probability that both will
have (i) different birthdays? (ii) the same birthday? (ignoring a leap year).
Solution : Out of the two friends, one girl, say, Savita’s birthday can be any day of the
year. Now, Hamida’s birthday can also be any day of 365 days in the year.
We assume that these 365 outcomes are equally likely.
(i) If Hamida’s birthday is different from Savita’s, the number of favourable outcomes
for her birthday is 365 – 1 = 364
So, P (Hamida’s birthday is different from Savita’s birthday) =
364
365
(ii) P(Savita and Hamida have the same birthday)
= 1 – P (both have different birthdays)
=
364
1
365
− [Using P( E ) = 1 – P(E)]
=
1
365
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PROBABILITY 303
Example 7 : There are 40 students in Class X of a school of whom 25 are girls and 15
are boys. The class teacher has to select one student as a class representative. She
writes the name of each student on a separate card, the cards being identical. Then
she puts cards in a bag and stirs them thoroughly. She then draws one card from the
bag. What is the probability that the name written on the card is the name of (i) a girl?
(ii) a boy?
Solution : There are 40 students, and only one name card has to be chosen.
(i) The number of all possible outcomes is 40
The number of outcomes favourable for a card with the name of a girl = 25 (Why?)
Therefore, P (card with name of a girl) = P(Girl) =
25 5
40 8
=
(ii) The number of outcomes favourable for a card with the name of a boy = 15 (Why?)
Therefore, P(card with name of a boy) = P(Boy) =
15 3
40 8
=
Note : We can also determine P(Boy), by taking
P(Boy) = 1 – P(not Boy) = 1 – P(Girl) =
5 3
1
8 8
− =
Example 8 : A box contains 3 blue, 2 white, and 4 red marbles. If a marble is drawn
at random from the box, what is the probability that it will be
(i) white? (ii) blue? (iii) red?
Solution : Saying that a marble is drawn at random is a short way of saying that all the
marbles are equally likely to be drawn. Therefore, the
number of possible outcomes = 3 +2 + 4 = 9 (Why?)
Let W denote the event ‘the marble is white’, B denote the event ‘the marble is blue’
and R denote the event ‘marble is red’.
(i) The number of outcomes favourable to the event W = 2
So, P(W) =
2
9
Similarly, (ii) P(B) =
3
9
=
1
3
and (iii) P(R) =
4
9
Note that P(W) + P(B) + P(R) = 1.
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304 MATHEMATICS
Example 9 : Harpreet tosses two different coins simultaneously (say, one is of `1
and other of ` 2). What is the probability that she gets at least one head?
Solution : We write H for ‘head’ and T for ‘tail’. When two coins are tossed
simultaneously, the possible outcomes are (H, H), (H, T), (T, H), (T, T), which are all
equally likely. Here (H, H) means head up on the first coin (say on ` 1) and head up
on the second coin (` 2). Similarly (H, T) means head up on the first coin and tail up on
the second coin and so on.
The outcomes favourable to the event E, ‘at least one head’ are (H, H), (H, T)
and (T, H). (Why?)
So, the number of outcomes favourable to E is 3.
Therefore, P(E) =
3
4
i.e., the probability that Harpreet gets at least one head is
3
4
⋅
Note : You can also find P(E) as follows:
P (E) =
1 3
1 – P(E) = 1 –
4 4
=
1
Since P(E) = P(no head) =
4
 
 
 
Did you observe that in all the examples discussed so far, the number of possible
outcomes in each experiment was finite? If not, check it now.
There are many experiments in which the outcome is any number between two
given numbers, or in which the outcome is every point within a circle or rectangle, etc.
Can you now count the number of all possible outcomes? As you know, this is not
possible since there are infinitely many numbers between two given numbers, or there
are infinitely many points within a circle. So, the definition of (theoretical) probability
which you have learnt so far cannot be applied in the present form. What is the way
out? To answer this, let us consider the following example :
Example 10* : In a musical chair game, the person playing the music has been
advised to stop playing the music at any time within 2 minutes after she starts playing.
What is the probability that the music will stop within the first half-minute after starting?
Solution : Here the possible outcomes are all the numbers between 0 and 2. This is
the portion of the number line from 0 to 2 (see Fig. 15.1).
Fig. 15.1
* Not from the examination point of view.
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PROBABILITY 305
Let E be the event that ‘the music is stopped within the first half-minute’.
The outcomes favourable to E are points on the number line from 0 to
1
2
.
The distance from 0 to 2 is 2, while the distance from 0 to
1
2
is
1
2
.
Since all the outcomes are equally likely, we can argue that, of the total distance
of 2, the distance favourable to the event E is
1
2
.
So, P(E) =
Distance favourable to the event E
Total distance in which outcomes can lie
=
1
12
2 4
=
Can we now extend the idea of Example 10 for finding the probability as the ratio of
the favourable area to the total area?
Example 11* : A missing helicopter is reported to have crashed somewhere in the
rectangular region shown in Fig. 15.2. What is the probability that it crashed inside the
lake shown in the figure?
Fig. 15.2
Solution : The helicopter is equally likely to crash anywhere in the region.
Area of the entire region where the helicopter can crash
= (4.5 × 9) km2
= 40.5 km2
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306 MATHEMATICS
Area of the lake = (2.5 × 3) km2
= 7.5 km2
Therefore, P (helicopter crashed in the lake) =
7.5 5
40.5 405 27
7 5
= =
Example 12 : A carton consists of 100 shirts of which 88 are good, 8 have minor
defects and 4 have major defects. Jimmy, a trader, will only accept the shirts which
are good, but Sujatha, another trader, will only reject the shirts which have major
defects. One shirt is drawn at random from the carton. What is the probability that
(i) it is acceptable to Jimmy?
(ii) it is acceptable to Sujatha?
Solution : One shirt is drawn at random from the carton of 100 shirts. Therefore,
there are 100 equally likely outcomes.
(i) The number of outcomes favourable (i.e., acceptable) to Jimmy = 88 (Why?)
Therefore, P (shirt is acceptable to Jimmy) =
88
0.88
100
=
(ii) The number of outcomes favourable to Sujatha = 88 + 8 = 96 (Why?)
So, P (shirt is acceptable to Sujatha) =
96
0.96
100
=
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
(i) 8? (ii) 13? (iii) less than or equal to 12?
Solution : When the blue die shows ‘1’, the grey die could show any one of the
numbers 1, 2, 3, 4, 5, 6. The same is true when the blue die shows ‘2’, ‘3’, ‘4’, ‘5’ or
‘6’. The possible outcomes of the experiment are listed in the table below; the first
number in each ordered pair is the number appearing on the blue die and the second
number is that on the grey die.
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PROBABILITY 307
1 2 3 4 5 6
1 (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
2 (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
3 (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
4 (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
5 (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
6 (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
Fig. 15.3
Note that the pair (1, 4) is different from (4, 1). (Why?)
So, the number of possible outcomes = 6 × 6 = 36.
(i) The outcomes favourable to the event ‘the sum of the two numbers is 8’ denoted
by E, are: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) (see Fig. 15.3)
i.e., the number of outcomes favourable to E = 5.
Hence, P(E) =
5
36
(ii) As you can see from Fig. 15.3, there is no outcome favourable to the event F,
‘the sum of two numbers is 13’.
So, P(F) =
0
0
36
=
(iii) As you can see from Fig. 15.3, all the outcomes are favourable to the event G,
‘sum of two numbers ≤ 12’.
So, P(G) =
36
1
36
=
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308 MATHEMATICS
EXERCISE 15.1
1. Complete the following statements:
(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 .
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.
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?
4. Which of the following cannot be the probability of an event?
(A)
2
3
(B) –1.5 (C) 15% (D) 0.7
5. If P(E) = 0.05, what is the probability of ‘not E’?
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?
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?
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?
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?
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PROBABILITY 309
10. A piggy bank contains hundred 50p coins, fifty ` 1 coins, twenty ` 2 coins and ten ` 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 ` 5 coin?
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?
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 Fig. 15.5 ), 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?
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.
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
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?
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.
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 ?
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.
Fig. 15.5
Fig. 15.4
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310 MATHEMATICS
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?
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. 15.6
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 ?
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.
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.
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?
[Hint : Throwing a die twice and throwing two dice simultaneously are treated as the
same experiment]
3m
2m
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PROBABILITY 311
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
.
EXERCISE 15.2 (Optional)*
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?
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:
+ 1 2 2 3 3 6
1 2 3 3 4 4 7
2 3 4 4 5 5 8
2 5
3
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?
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.
4. A box contains 12 balls out of which x are black. If 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.
* These exercises are not from the examination point of view.
Number in first throw
Numberinsecondthrow
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312 MATHEMATICS
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
2
3
⋅ Find the number of blue balls
in the jar.
15.3 Summary
In this chapter, you have studied the following points :
1. The difference between experimental probability and theoretical probability.
2. The theoretical (classical) probability of an event E, written as P(E), is defined as
P(E)=
Number of all possible outcomes of the experiment
where we assume that the outcomes of the experiment are equally likely.
3. The probability of a sure event (or certain event) is 1.
4. The probability of an impossible event is 0.
5. The probability of an event E is a number P(E) such that
0≤P(E)≤1
6. An event having only one outcome is called an elementary event. The sum of the
probabilities of all the elementary events of an experiment is 1.
7. For any event E, P (E) + P ( E ) = 1, where E stands for ‘not E’. E and E are called
complementary events.
ANOTE TO THE READER
The experimental or empirical probability of an event is based on
what has actually happened while the theoretical probability of the
event attempts to predict what will happen on the basis of certain
assumptions. As the number of trials in an experiment, go on
increasing we may expect the experimental and theoretical
probabilities to be nearly the same.
2019-20

Ncert class-10-mathematics-chapter-15

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