## 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.**

**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) = \(\frac { 285 }{ 500 } \) = \(\frac { 57 }{ 100 } \) = 0.57

(ii) P(E2) = \(\frac { 215 }{ 500 } \) = \(\frac { 43 }{ 100 } \) = 0.43

**Question 2.**

**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) = \(\frac { 112 }{ 400 } \) = \(\frac { 28 }{ 100 } \) = 0.28

(ii)P(E2) = \(\frac { 160 }{ 400 } \) = \(\frac { 40 }{ 100 } \)= 0.40

(iii) P(E3) = \(\frac { 128 }{ 400 } \) = \(\frac { 32 }{ 100 } \) = 0.32 Ans.

**Question 3.**

**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) = \(\frac { 39 }{ 200 } \) = 0.195

(ii) P(E3) = \(\frac { 67 }{ 200 } \) = 0.335

(iii) P(E4) = \(\frac { 36 }{ 200 } \) = \(\frac { 18 }{ 100 } \) = 0.18

(iv) P(E2) = \(\frac { 58 }{ 200 } \) = \(\frac { 29 }{ 100 } \) = 0.29

**Question 4.**

**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) = \(\frac { 54 }{ 300 } \) = \(\frac { 18 }{ 100 } \) = 0.18

(ii) P(E6) = \(\frac { 33 }{ 100 } \) = \(\frac { 11 }{ 100 } \)= 0.11

(iii) P(E5) = \(\frac { 39 }{ 300 } \) = \(\frac { 13 }{ 100 } \) = 0.13

(iv) P(E1) = \(\frac { 60 }{ 300 } \) = \(\frac { 20 }{ 100 } \)= 0.20 Ans.

**Question 5.**

**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) = \(\frac { 142 }{ 200 } \) = \(\frac { 71 }{ 100 } \) = 0.71

(ii) P(E2) = \(\frac { 58 }{ 200 } \) = \(\frac { 29 }{ 100 } \) = 0.29 Ans.

**Question 6.**

**Solution:**

Total number of tests = 6

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

Probability will he

P(E) = \(\frac { 2 }{ 6 } \) = \(\frac { 1 }{ 3 } \)Ans.

**Question 7.**

**Solution:**

No. of vehicles of various types = 240

No. of vehicles of two wheelers = 64.

Probability will be P(E) = \(\frac { 84 }{ 240 } \) = \(\frac { 7 }{ 20 } \) = 0.35 Ans.

**Question 8.**

**Solution:**

No. of phone numbers are one page = 200

Let E be the number of events in each case,

Then (i) P(E5) = \(\frac { 24 }{ 200 } \) = \(\frac { 12 }{ 100 } \) = 0.12

(ii) P(E8) = \(\frac { 16 }{ 200 } \) = \(\frac { 8 }{ 100 } \) = 0.08 Ans.

**Question 9.**

**Solution:**

No. of students whose blood group is checked = 40

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

Then (i) P(E0) = \(\frac { 14 }{ 40 } \) = \(\frac { 7 }{ 20 } \) = 0.35

(ii) P(EAB) = \(\frac { 6 }{ 40 } \) = \(\frac { 3 }{ 20 } \) = 0.15 Ans.

**Question 10.**

**Solution:**

No. of total students = 30.

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

P(E) = \(\frac { 6 }{ 30 } \) = \(\frac { 1 }{ 5 } \) = 0.2 Ans.

**Question 11.**

**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) = \(\frac { 60 }{ 360 } \) = \(\frac { 1 }{ 6 } \)

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

P(E) = \(\frac { 80 }{ 360 } \) = \(\frac { 2 }{ 9 } \)

(iii) Less than 10 years = zero

P(E) = \(\frac { 0 }{ 360 } \) = 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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