## RS Aggarwal Class 9 Solutions Chapter 15 Probability Ex 15A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 15 Probability Ex 15A.

**Question 1.**

**A coin is tossed 500 times and we get head : 285 times, tail : 215 times.When a coin is tossed at random, what is the probability of getting**

**(i) a head ? (ii) a tail?**

**Solution:**

Number of trials = 500 times

Let E be the no. of events in each case, then

∴No. of heads (E1) = 285 times

and no. of tails (E2) = 215 times

∴ Probability in each case will be

∴(i)P(E1) = = = 0.57

(ii) P(E2) = = = 0.43

**Question 2.**

**Two coins are tossed 400 times and we get two heads: 112 times ; one head; 160 times; 0 head: 128 times.**

**When two coins are tossed at random, what is the probability of getting**

**(i) 2 heads ? (ii) 1 head? (iii) 0 head?**

**Solution:**

No. of trials = 400

Let E be the no. of events in each case, then

No. of 2 heads (E1) = 112

No. of one head (E2) = 160 times

and no. of O. head (E3) = 128 times

∴ Probability in each case will be:

∴ (i)P(E1) = = = 0.28

(ii)P(E2) = = = 0.40

(iii) P(E3) = = = 0.32 Ans.

**Question 3.**

**Three coins are tossed 200 times and we get three heads: 39 times ; two heads: 58 times; one head : 67 times ; 0 head : 36 times.**

**When three coins are tossed at random, what is the probability of getting**

**(i) 3 heads ? (ii) 1 head? (iii) 0 head ? (iv) 2 heads ?**

**Solution:**

Number of total trials = 200

Let E be the no. of events in each case, then

No. of three heads (E1) = 39 times

No. of two heads (E2) = 58 times

No. of one head (E3) = 67 times

and no. of no head (E4) = 36 times

∴ Probability in each case will be .

(i) P(E1) = = 0.195

(ii) P(E3) = = 0.335

(iii) P(E4) = = = 0.18

(iv) P(E2) = = = 0.29

**Question 4.**

**A die is thrown 300 times and the outcomes are noted as given below:**

**When a die is thrown at random, what is the probability of getting a**

**(i) 3? (ii) 6? (iii) 5? (iv) 1?**

**Solution:**

Solution No. of trials = 300 times

Let E be the no. of events in each case, then

No. of outcome of 1(E1) = 60

No. of outcome of 2(E2) = 72

No. of outcome of 3(E3) = 54

No. of outcome of 4(E4) 42

No. of outcome of 5(E5) = 39

No. of outcome of 6(E6) = 33

The probability of

(i) P(E3) = = = 0.18

(ii) P(E6) = = = 0.11

(iii) P(E5) = = = 0.13

(iv) P(E1) = = = 0.20 Ans.

**Question 5.**

**In a survey of 200 ladies, it was found that 142 like coffee, while 58 dislike it.**

**Find the probability that a lady chosen at random**

**(i) likes coffee, (ii) dislikes coffee.**

**Solution:**

No. of ladies on whom survey was made = 200.

Let E be the no. of events in each case.

No. of ladies who like coffee (E1) = 142

No. of ladies who like coffee (E2) = 58

Probability of

(1) P(E1) = = = 0.71

(ii) P(E2) = = = 0.29 Ans.

**Question 6.**

**The percentages of marks obtained by a student in six unit tests are given below :**

**A unit test is selected at random. What is the probability that the student gets more than 60% marks in the test ?**

**Solution:**

Total number of tests = 6

No. of test in which the students get more than 60% mark = 2

Probability will he

P(E) = = Ans.

**Question 7.**

**On a particular day, at a crossing in a city, the various types of 240 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 a two-wheeler ?**

**Solution:**

No. of vehicles of various types = 240

No. of vehicles of two wheelers = 64.

Probability will be P(E) = = = 0.35 Ans.

**Question 8.**

**On one page of a telephone directory, there are 200 phone numbers. The frequency distribution of their units digits is given below :**

**One of the numbers is chosen at random from the page. What is the probability that the units digit of the chosen number is (i) 5 ? (ii) 8 ?**

**Solution:**

No. of phone numbers are one page = 200

Let E be the number of events in each case,

Then (i) P(E5) = = = 0.12

(ii) P(E8) = = = 0.08 Ans.

**Question 9.**

**The following table shows the blood groups of 40 students of a class.**

**One student of the class is chosen at random. What is the probability that the chosen student has blood group (i) O ? (ii) AB ?**

**Solution:**

No. of students whose blood group is checked = 40

Let E be the no. of events in each case,

Then (i) P(E0) = = = 0.35

(ii) P(EAB) = = = 0.15 Ans.

**Question 10.**

**The table given below shows the marks obtained by 30 students in a test.**

**Out of these students, one is chosen at random. What is the probability that his marks lie in the interval of 21 – 30 ?**

**Solution:**

No. of total students = 30.

Let E be the number of elements, this probability will be of interval 21 – 30

P(E) = = = 0.2 Ans.

**Question 11.**

**Following are the ages (in years) of 360 patients, getting medical treatment in a hospital:**

**One of the patients is selected at random.**

**Find the probability that his age is**

**(i) 30 years or more but less than 40 years,**

**(ii) 50 years or more but less than 70 years,**

**(iii) less than 10 years,**

**(iv) 10 years or more.**

**Solution:**

Total number of patients of various age group getting medical treatment = 360

Let E be the number of events, then

(i) No. of patient which are 30 years or more but less than 40 years = 60.

P(E) = =

(ii) 50 years or more but less than 70 years = 50 + 30 = 80

P(E) = =

(iii) Less than 10 years = zero

P(E) = = 0

(iv) 10 years or more 90 + 50 + 60 + 80 + 50 + 30 = 360

Hope given RS Aggarwal Class 9 Solutions Chapter 15 Probability Ex 15A are helpful to complete your math homework.

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