## 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:
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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