ICSE Class 10 Maths Chapter 25 Probability

25 Probability

25.1 Introduction

Consciously or unconsciously, all of us sometime use the phrases like: most likely, almost uncertain, most probably, no chance at all, etc.

If we read these phrases carefully, we will find that all of them involve an element of uncertainty. The measure of uncertainty is called the theory of probability.

The study of the theory of probabilities is of great mathematical interest and of great practical importance.

25.2 Some Basic Terms and Concepts

1. Experiment

A process which results in some well-defined outcome is known as an experiment.

For example:

(i) When a coin is tossed, we shall be getting either a head or a tail i.e. its outcome is a head or a tail, which is well-defined.

(ii) When a die is thrown the possible outcomes are 1, 2, 3, 4, 5 and 6, which are also well-defined.

2. Random Experiment

Random experiment means all the outcomes of the experiment are known in advance, but any specific outcome of the experiment is not known in advance.

For example:

(i) Tossing a coin is a random experiment because there are only two possible outcomes, head and tail, and these outcomes are known well in advance. But the specific outcome of the experiment i.e. whether a head or a tail is not known in advance.

(ii) Throwing a die is a random experiment because we know in advance that there are only six possible outcomes of the experiment i.e. 1, 2, 3, 4, 5 and 6. But it is not possible to know which of these six numbers will finally be the result.

A random experiment may result in two or more outcomes; for example: (i) tossing a coin. (ii) throwing a die, etc.

3. Sample Space

The set of all possible outcomes of an experiment is called sample space and is, in general, denoted by letter S.

For example:

(i) When we toss a coin once, it may come up in either of two ways: Head (H) or Tail (T). So, there are two possible outcomes of this random experiment. Thus the sample space (S) of this random experiment is given by S = {H, T}

(ii) When we roll a die once, it can land with any of its 6 faces pointing upward. Thus, the outcome of this experiment is getting any of the six numbers 1, 2, 3, 4, 5 and 6. Hence, the sample space for the experiment is S = {1, 2, 3, 4, 5, 6}.

(iii) When two coins are tossed together, the random experiment may result:

(a) head (H) on the first coin and head (H) on the second coin.

(b) head (H) on the first coin and tail (T) on the second coin.

(c) tail (T) on the first coin and head (H) on the second coin.

(d) tail (T) on the first coin and tail (T) on the second coin.

Thus the corresponding sample space S = {(H, H), (H, T), (T, H), (T, T)}.

(iv) When two dice are rolled together, the corresponding sample space for the random experiment is as given below:

S = {(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)}.

(v) When a coin and a die are tossed together, the corresponding sample space for the random experiment is as given below:

S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6),

(T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}.

4. Equally Likely Outcomes

In case of tossing a coin:

(i) It is known, in advance, that the coin will land with its head or tail up.

(ii) It is reasonably assumed that each outcome, a head or a tail, is as likely to occur as the other. In other words, we say that there are equal chances for the coin to land with its head or tail up.

Referring to the terms used in this chapter, we say that the outcomes, head and tail, are equally likely.

In case of throwing a die:

(i) It is known, in advance, that the dice will show the number 1, 2, 3, 4, 5 or 6 on the upper-most face.

(ii) It can reasonably be accepted that each of the numbers 1, 2, 3, 4, 5 and 6 has the same possibility to come to the upper-most face.

Hence, showing up the numbers 1, 2, 3, 4, 5 and 6 on the throwing of a die are equally likely outcomes.

Are the outcomes of all experiments equally likely? Suppose a bag contains 6 red and 2 yellow balls. Let a ball be drawn from the bag without looking into it. The ball that will come out will either be a red or a yellow ball.

Are the outcomes, a red ball and a yellow ball, equally likely? No.

Reason: Since the bag contains 6 red and 2 yellow balls, then in a single draw of a ball from this bag (without looking into it); it is more likely to get a red ball than a yellow ball. Hence, the outcomes are not equally likely.

However, if the bag contains equal numbers of red balls and yellow balls, the outcomes are equally likely.

5. An Event

An outcome of a random experiment is called an event. In other words, an event is something that happens.

On tossing a coin, the possible outcome is a head (H) or a tail (T). Here, getting a head or a tail is an event of the experiment of tossing a coin. Similarly, in throwing a cubical dice the six possible outcomes are 1, 2, 3, 4, 5 or 6. Thus, getting 1, 2, 3, 4, 5 or 6 on the upper face of the dice is an event of the experiment of throwing a dice.

In the same way, if a card is drawn from a well-shuffled pack of 52 playing cards, any one of them can be the outcome. So, there are 52 events of the random experiment of drawing a card from a pack of 52 playing cards.

25.3 Measurement of Probability

The probability of an event denotes the likelihood of its happening.

If in a random experiment, the total number of events (outcomes) are n out of which m events (outcomes) are favourable to a particular event E; then the probability of happening of event E is denoted by P(E) and is equal to the ratio \(\frac{m}{n}\).

i.e. P(E) = Probability of the happening of event E

\[= \frac{m}{n}\]

\[= \frac{\text{Number of events (outcomes) favourable to E}}{\text{Total number of all possible outcomes}}\]

For example:

If a die is rolled once, and an even number is required on the upper face of it; then in this experiment:

Total number of outcomes = 6 (any of 1, 2, 3, 4, 5 and 6) and, number of favourable outcomes = 3 (any of 2, 4 and 6)

\[\therefore \text{Probability of getting an even number on the upper face} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{3}{6} = \frac{1}{2}\]

1. Empirical (or, experimental) Probability

When the probability is based on an actual experiment, it is called an empirical (or, experimental) probability.

For example:

If a coin is tossed 100 times and the outcomes of this experiment are 57 heads and 43 tails, the probability of a head is \(\frac{57}{100}\) and that of a tail is \(\frac{43}{100}\). Since these probabilities are based on the actual experiment of tossing a coin 100 times; they are experimental (or, empirical) probabilities.

(i) For finding the experimental probability, an adequate recording of the outcomes is required.

(ii) Experimental probabilities are only 'estimates'. If the same experiment of tossing a coin 100 times is performed again, it will not necessarily give the same results of getting the number of heads and the number of tails. And, so the probabilities for a head and a tail will also not be the same.

2. Classical (or, theoretical) Probability

When a repetition of an experiment can be avoided for calculating the exact probability, the probability so obtained is called classical (or, theoretical) probability.

(i) The empirical probability can be applied to every event associated with an experiment which can be repeated a large number of times.

(ii) In theoretical (classical) probability, we make certain assumptions and one of these assumptions is that the outcomes are equally likely.

(iii) Probability of an event (outcome) = \(\frac{\text{Number of favourable outcomes}}{\text{Number of all possible outcomes}}\)

(iv) In this chapter, the probability means theoretical (or, classical) probability.

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

Solution:

In the random experiment of tossing a coin once, the total number of possible outcomes is 2 which are Head (H) and Tail (T).

Favourable outcome is 'getting a head'.

\[\Rightarrow \text{Number of favourable outcome} = 1\]

\[\therefore \text{P(getting a head)} = \frac{\text{Number of favourable outcomes}}{\text{Number of all possible outcomes}} = \frac{1}{2} \quad \text{Ans.}\]

An event, having only one favourable outcome, is called an elementary event, as shown in the example given above. Similarly, events in the following examples are also elementary events.

A bag contains a black ball, a red ball and a green ball, all the balls are identical in shape and size. Mohit takes out a ball from the bag, without looking into it. What is the probability that the ball drawn is: (i) red ball? (ii) black ball? (iii) green ball?

Solution:

When Mohit takes out a ball without looking into the bag, the outcomes of the experiment are equally likely.

Clearly, the total number of possible outcomes = 3

(i) The number of favourable outcome (getting a red ball) = 1

\[\Rightarrow \text{The probability of drawing a red ball} = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{1}{3} \quad \text{Ans.}\]

In short; P(red ball) = \(\frac{1}{3}\) Ans.

Similarly, (ii) P(drawing a black ball) = \(\frac{1}{3}\) Ans.

and, (iii) P(getting a green ball) = \(\frac{1}{3}\) Ans.

The sum of the probabilities of all the elementary events of an experiment is always one. In the example, given above:

P(red ball) + P(black ball) + P(green ball) = \(\frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1\)

In a single throw of a die, find the probability of getting a number: (i) greater than 2 (ii) less than or equal to 2 (iii) not greater than 2.

Solution:

In a single throw of a die, the total possible outcomes are 6 (1, 2, 3, 4, 5 and 6).

(i) Out of 1, 2, 3, 4, 5 and 6; the numbers greater than 2 are 3, 4, 5 and 6.

\[\therefore \text{Total number of favourable outcomes} = 4 \text{ (3, 4, 5 and 6)}\]

\[\Rightarrow \text{P(getting a number greater than 2)} = \frac{\text{Number of favourable outcomes}}{\text{Number of all possible outcomes}} = \frac{4}{6} = \frac{2}{3} \quad \text{Ans.}\]

(ii) Out of all possible outcomes 1, 2, 3, 4, 5 and 6, the numbers less than or equal to 2 are 1 and 2.

\[\therefore \text{Total number of favourable outcomes} = 2\]

\[\Rightarrow \text{P(getting a number less than or equal to 2)} = \frac{2}{6} = \frac{1}{3} \quad \text{Ans.}\]

(iii) Out of all possible outcomes 1, 2, 3, 4, 5 and 6, the numbers not greater than 2 are 1 and 2 only.

\[\therefore \text{The number of favourable outcomes} = 2\]

\[\Rightarrow \text{P(getting a number not greater than 2)} = \frac{2}{6} = \frac{1}{3} \quad \text{Ans.}\]

1. In a single throw of a die; getting a number less than or equal to 2 and getting a number not greater than 2 mean the same.

For this reason: P(getting a number less than or equal to 2) = P(getting a number not greater than 2)

2. If the event of getting number greater than 2 is denoted by E. Then the event of getting a number not greater than 2 (or, a number less than or equal to 2) is denoted by not E or by \(\overline{E}\).

Thus, \[\text{P(E)} = \text{P(getting number greater than 2)} = \frac{2}{3}\]

And, \[\text{P}\left(\overline{\text{E}}\right) = \text{P(not E)} = \text{P(getting number not greater than 2)} = \frac{1}{3}\]

\[\therefore \text{P(E)} + \text{P}\left(\overline{\text{E}}\right) = \frac{2}{3} + \frac{1}{3} = 1\]

Also, \[\text{P(E)} + \text{P}\left(\overline{\text{E}}\right) = 1 \Rightarrow \text{P}\left(\overline{\text{E}}\right) = 1 - \text{P(E)}\]

i.e. \[\text{P(not E)} = 1 - \text{P(E)}\]

3. E and \(\overline{E}\) (not E) are called complementary events i.e., for any event E, the event of non-occurrence of E is called its complementary event and is denoted by \(\overline{E}\).

4. The sum of probabilities of an event and its complementary event is always 1.

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