ICSE Class 10 Maths Chapter 30 Probability

Chapter 30: Probability

Points To Remember

1. Probability: It is the concept which numerically measures the degree of certainty of the occurence of an event.

2. Experiment: An operation which can be produce some well define outcomes, is called an experiment.

3. Event: Each outcome of an experiment is called an event.

4. Trial: By a trial, we mean performing an experiment.

5. Empirical Probability: Suppose we perform an experiment and let n be the total number of trials. The empirical probability of the happening of an event E is defined as:

\[P(E) = \frac{\text{Number of trials in which the event happened}}{\text{Total number of trials}}\]

Classical Probability

(a) Sample Space: In a random experiment, the set of all possible outcomes is called a sample space and it is denoted by S.

(b) Event: The collection of some or all outcomes in an experiment is called an event. If F is an event, then clearly E \(\subseteq\) S.

Probability Of Occurence Of An Event (Classical Definition)

In an random experiment, let S be the sample space and E be the event, then E \(\leq\) S.

The probability of occurence of E, is defined as:

\[P(E) = \frac{\text{Number of distinct elements in E}}{\text{Number of distinct elements in S}} = \frac{n(E)}{n(S)}\]

Tossing a coin, throwing a die, drawing cards from a well-shuffled pack of 52 cards etc. are the examples of classical probability.

(a) Sure Event: The probability of a sure event is 1

(b) Impossible Event: The probability of an impossible event is 0.

(c) Complementary Event: Let E be an event and (not E) be an event which occurs only when E does not occur. The event (not E) is called the complementary event of E

\[P(E) + P(\text{not } E) = 1\]

\[\Rightarrow P(E) = 1 - P(\text{not } E)\]

Note: (i) P(not E) can also be written as P(E')

(ii) For any event E, we have \(0 \leq P(E) \leq 1\)

Teacher's Note

Probability helps us understand the likelihood of everyday events like weather forecasts, sports outcomes, and medical test results. Understanding these concepts makes us better decision-makers in uncertain situations.

Exercise 30 (A)

Question 1

A coin is tossed 200 times and it was found that head appears 72 times and tail appears 128 times.

If a coin is tossed at random what is the probability of getting (i) a head (ii) a tail?

Solution

A coin is tossed 200 times.

Number of heads appear = 72

Number of tails appears = 128

(i) \[P(E_1) = \frac{\text{Number of heads}}{\text{Total number of trials}} = \frac{72}{200} = \frac{9}{25}\]

(ii) \[P(E_2) = \frac{\text{Number of tails}}{\text{Total number of trials}} = \frac{128}{200} = \frac{16}{25} \text{ Ans.}\]

Question 2

Two coins are tossed simultaneously 125 times and it was observed that both heads appeared 15 times. If two coins are tossed simultaneously at random. What is the probability of getting both heads?

Solution

Total number of times two coins tossed simultaneously = 125

Number of both heads appeared = 15

\[P(E) = \frac{\text{Number of heads}}{\text{Total number of trials}} = \frac{15}{125} = \frac{3}{25} \text{ Ans.}\]

Question 3

A die is thrown 260 times. Prime numbers appear on the upper face 39 times. If a die is thrown at random, what is the probability of getting a prime number?

Solution

A die is thrown 260 times

Prime number on the faces of a die are 2, 3, 5

Number of prime numbers which appears on the face = 39 times

\[P(E) = \frac{39}{260} = \frac{3}{20} \text{ Ans.}\]

Question 4

A survey of 650 men, showed that only 52 of them know English. Out of these men, if one is selected at random, what is the probability that the selected man knows English?

Solution

Total number of men = 650

Number of men who know English = 52

Probability will be

\[P(E) = \frac{52}{650} = \frac{2}{25} \text{ Ans.}\]

Question 5

On the particular day, at a crossing in a city, the various types of 280 vehicles going past during a time-interval, were observed as under.

Out of these vehicles, one is chosen at random. What is the probability that the chosen vehicle is (i) a four-wheeler, (ii) a two-wheeler, (iii) a three-wheeler?

Solution

Total number of wheelers = 91 + 63 + 126 = 280

(i) Now probability of four wheeler

\[P(E_1) = \frac{\text{Number of four - wheelers}}{\text{Total number of wheelers}} = \frac{126}{280} = \frac{9}{20}\]

(ii) Probability of two wheeler

\[P(E_2) = \frac{\text{Number of two wheelers}}{\text{Total number of wheelers}} = \frac{91}{280} \text{ Ans.}\]

(iii) Probability of three wheelers

\[P(E_3) = \frac{\text{Number of three wheelers}}{\text{Total number of wheelers}} = \frac{63}{280} = \frac{9}{40} \text{ Ans.}\]

Question 6

The following table shows the blood-groups of 60 students of a class:

One student of the class is chosen at random. What is the probability that the chosen student has blood groups. (i) O (ii) AB (iii) A (iv) B?

Solution

Number of total students in a class = 16 + 12 + 23 + 9 = 60

(i) Probability of students having blood group O

\[P(E_1) = \frac{\text{Number of students having blood group O}}{\text{Total number of students}} = \frac{23}{60} \text{ Ans.}\]

(ii) Probability of students having blood group AB

\[P(E_2) = \frac{\text{Number of students having blood group AB}}{\text{Total number of students}} = \frac{9}{60} = \frac{3}{20} \text{ Ans.}\]

(iii) Probability of students having blood group A

\[P(E_3) = \frac{\text{Number of students having blood group A}}{\text{Total number of students}} = \frac{16}{60} = \frac{4}{15} \text{ Ans.}\]

(iv) Probability of students having blood group B

\[P(E_4) = \frac{\text{Number of students having blood group B}}{\text{Total number of students}} = \frac{12}{60} = \frac{1}{5} \text{ Ans.}\]

Teacher's Note

Real-world data like blood type distributions helps us understand human genetic diversity and the importance of probability in medical contexts, where knowing blood type frequencies can be life-saving during emergencies.

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