Please Wait... While Loading Full Video#### Class-10 chapter -15

### PROBABILITY

`color{olive}( "Introduction ")`

`color{olive}("Probability : A Theoretical Approach")`

`color{olive}("Probability : A Theoretical Approach")`

As we've 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.

`"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.

You tossed a coin many times and noted the number of times it turned up heads (or tails). 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 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 `5067/10000 = 0.5067` 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 up to, 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.

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.

`"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.

You tossed a coin many times and noted the number of times it turned up heads (or tails). 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 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 `5067/10000 = 0.5067` 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 up to, 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.

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.

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) = text(Number of trials in which the event happened)/text(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 multistoreyed 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) = text(Number of outcomes favourable to E)/text(Number of all possible outcomes of the experiment)`

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.

Let us find the probability for some of the events associated with experiments where the equally likely assumption holds.

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.

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) = text(Number of trials in which the event happened)/text(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 multistoreyed 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) = text(Number of outcomes favourable to E)/text(Number of all possible outcomes of the experiment)`

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.

Let us find the probability for some of the events associated with experiments where the equally likely assumption holds.

Q 3270001816

Find the probability of getting a head when a coin is tossed once. Also find the probability of getting a tail.

Class 10 Chapter 15 Example 1

Class 10 Chapter 15 Example 1

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(text(head)) = text(Number of outcomes favourable to E)/text(Number of all possible outcomes) = 1/2`

Similarly, if F is the event ‘getting a tail’, then

`P(F) = P(text(tail)) = 1/2`

Q 3280001817

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?

Class 10 Chapter 15 Example 2

(i) yellow ball? (ii) red ball? (iii) blue ball?

Class 10 Chapter 15 Example 2

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`

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.

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.

Q 3210001819

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 ?

Class 10 Chapter 15 Example 3

Class 10 Chapter 15 Example 3

(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`

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.

From Example 1, we note that

`P(E) + P(F) = 1/2+1/2 = 1` ..........(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) = 1/3+2/3 = 1` ............(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 `barE` .

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(barE ) = 1 – P(E)`

The event `barE` , representing ‘not `E`’, is called the complement of the event `E`. We also say that `E` and `barE` 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`

That is, the probability of an event which is impossible to occur is 0. Such an event is called an impossible event.

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 (♠), hearts (), diamonds (⧫) and clubs (¨♣). 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.

`P(E) + P(F) = 1/2+1/2 = 1` ..........(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) = 1/3+2/3 = 1` ............(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 `barE` .

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(barE ) = 1 – P(E)`

The event `barE` , representing ‘not `E`’, is called the complement of the event `E`. We also say that `E` and `barE` 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`

That is, the probability of an event which is impossible to occur is 0. Such an event is called an impossible event.

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 (♠), hearts (), diamonds (⧫) and clubs (¨♣). 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.

Q 3210101910

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.

Class 10 Chapter 15 Example 4

(i) be an ace,

(ii) not be an ace.

Class 10 Chapter 15 Example 4

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`

Therefore, `P(E) = 4/52 = 1/13`

(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?)

The number of possible outcomes `= 52`

Therefore, `P(F) = 48/52 = 12/13`

Remark : Note that `F` is nothing but `barE` . Therefore, we can also calculate `P(F)` as

follows: `P(F) = P(bar E ) = 1 – P(E) = 1-1/13 = 12/13`

Q 3220101911

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?

Class 10 Chapter 15 Example 5

Class 10 Chapter 15 Example 5

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`

Q 3230101912

Savita and Hamida are friends. What is the probability that both will have (i) different birthdays? (ii) the same birthday? (ignoring a leap year).

Class 10 Chapter 15 Example 6

Class 10 Chapter 15 Example 6

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 (text(both have different birthdays))`

`= 1-364/365` [Using `P( barE ) = 1 – P(E)`]

` = 1/365`

Q 3240101913

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?

Class 10 Chapter 15 Example 7

Class 10 Chapter 15 Example 7

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

Therefore, P (card with name of a girl) = P(Girl) ` = 25/40 = 5/8`

ii) The number of outcomes favourable for a card with the name of a boy = 15

Therefore, P(card with name of a boy) = P(Boy) `= 15/40 = 3/8`

Note : We can also determine P(Boy), by taking

P(Boy) = 1 – P(not Boy) = 1 – P(Girl) `= 1-5/8 = 3/8`

Q 3250101914

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?

Class 10 Chapter 15 Example 8

(i) white? (ii) blue? (iii) red?

Class 10 Chapter 15 Example 8

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.`

Q 3220612511

Harpreet tosses two different coins simultaneously (say, one is of Re 1 and other of Rs 2). What is the probability that she gets at least one head?

Class 10 Chapter 15 Example 9

Class 10 Chapter 15 Example 9

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 Re 1) and head up on the second coin (Rs 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- P(barE) = 1-1/4 = 3/4` ( since `P(barE) = P (text(no head)) = 1/4`)

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 :

Q 3240612513

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?

Class 10 Chapter 15 Example 10

Class 10 Chapter 15 Example 10

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).

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) = text(Distance favourable to the event E)/text(Total distance in which outcomes can lie) = (1/2)/(2) = 1/4`

Can we now extend the idea of Example 10 for finding the probability as the ratio of

the favourable area to the total area?

Q 3260612515

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?

Class 10 Chapter 15 Example 11

Class 10 Chapter 15 Example 11

The helicopter is equally likely to crash anywhere in the region.

Area of the entire region where the helicopter can crash

`= (4.5 × 9) km^2 = 40.5 km^2`

Area of the lake `= (2.5 × 3) km^2 = 7.5 km^2`

Therefore, P (helicopter crashed in the lake) `(7.5)/(40.5) = 75/405 = 5/27`

Q 3270612516

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?

Class 10 Chapter 15 Example 12

(i) it is acceptable to Jimmy?

(ii) it is acceptable to Sujatha?

Class 10 Chapter 15 Example 12

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) ` = 88/100 = 0.88`

(ii) The number of outcomes favourable to Sujatha `= 88 + 8 = 96 ` (Why?)

So, P (shirt is acceptable to Sujatha) ` = 96/100 = 0.96`

Q 3200612518

Two dice, one blue and one grey, are thrown at the same time. Write own 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?

Class 10 Chapter 15 Example 13

(i) 8? (ii) 13? (iii) less than or equal to 12?

Class 10 Chapter 15 Example 13

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.

Note that the pair (1, 4) is different from (4, 1).

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) = 0/36 = 0`

(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) = 36/36 = 1`