Samacheer Kalvi 10th Maths Solutions Chapter 8 Exercise 8.4 Pdf

Samacheer Kalvi 10th Maths Solutions Chapter 8 Exercise 8.4 Pdf

TN Samacheer Kalvi 10th Maths Solutions Chapter 8 Exercise 8.4: Hello students, Are You looking for Samacheer Kalvi 10th Maths Solutions Chapter 8 Exercise 8.4 – Statistics and Probability on internet. If yes, here we have given TNSCERT Class 10 Maths Solution Chapter 8 Exercise 8.4 Pdf Guide for Samacheer Kalvi 10th Mathematics Solution for Exercise 8.4

TN Samacheer Kalvi 10th Maths Solutions Chapter 8 – Statistics and Probability

Board TNSCERT Class 10th Maths
Class 10
Subject Maths
Chapter 8 (Exercise 8.4)
Chapter Name Statistics and Probability

TNSCERT Class 10th Maths Pdf | all Exercise Solution

 

Exercise – 8.4

 

(1) If P(A) = 2/3, P(B) = 2/5, P(A∪B) = 1/3 then find P (A∪B).

Solution:

Given that, P(A) = 2/3, P(B) = 2/5, P (A∪B) = 2/3

We know that P(A∪B) = P(A) + P(B) – P(A∩B)

1/3 = 2/3 + 2/5 – P (A∩B)

P(A∩B) = 2/3 + 2/5 – 1/3

P (A∩B) = 10+6-5/15

P (A∩B) = 4/15

Thus, the P (A∩B) = 11/15

 

(2) A and B are two events such that, P(A) = 0.42, P(B) = 0.48, and P(A∩B) = 0.16. Find

(i) P (not A)

(ii) P (not B)

(iii) P (A or B)

Solution:

(i) P (not A) = P (A) and given P(A) = 0.42

∴ We know that P(A) + P(A) = 1

P(A) = 1 – P(A) = 1-0.42 = 0.58

Therefore, the value of P (not A) is 0.58

 

(iii) P (not B) = P (B) and given P(B) = 0.48

We know that P(B) + P(B) = 1

P(B) = 1 – P(B)

= 1 – 0.48

= 0.52

Therefore the value of P (not B) is 0.52

 

(iii) P (A or B) = P (A∪B)

Given, P(A) = 0.42, P (B) = 0.48 and P(A∩B) = 0.14

We know that, P(A∪B) = P(A) + P(B) – P(A∩B)

= 0.42 + 0.48 – 0.16

= 0.90 – 0.16

= 0.74

Therefore the value of P (A or B) is 0.74

 

(3) If A and B are two mutually exclusive events of a random experiment and P (not A) = 0.45, P (A∪B) = 0.65, then find P(B).

Solution:

Given, P (not A) = 0.45 and P(A∪B) = 0.65

P (A) = 0.45

∴ P (A) = 1 – P(A) = 1-0.45

= 0.55

Now, A and B are two mutually exclusive events of a random experiment.

∴ P (A∩B) = ⌀

Now, P (A∪B) = P(A) + P(B)

0.65 = 0.55 + P(B)

P(B) = 0.10

Thus, the value of P(B) is 0.10

 

(4) The probability that at least one of A and B occur is 0.6. If A and B occur simultaneously with probability 0.2, then find P(A) + P(B).

Solution:

Given that, P (A∪B) = 0.6 and P (A∩B) = 0.2

Now, P (A∪B) = P(A)  + P(B) – P (A∩B)

0.6 = P(A) + P(B) – 0.2

P(A) + P(B) = 0.6 + 0.2

1-P (A) + 1-P (B) = 0.8

2 – P(A) – P(B) = 0.8

[P(A) = 1 – P(A), P(B) = 1 – P(B)]

P(A) + P(B) = 2 – 0.8

P(A) + P(B) = 1.2

Thus, the value of P(A) + P(B) is 1.2

 

(5) The probability of happening of an event A is 0.5 and that of B is 0.3. If A and B are mutually exclusive events, then find the probability that neither A nor B happen.

Solution:

Given that, P(A) = 0.5 and P(B) = 0.3

Now, A and B are mutually exclusive events.

∴ P (A∩B) = ⌀

Now, P(A) ∪ P(B) = 1 {P(A) + P(B)}

= 1 – {0.5 + 0.3}

= 1 – 0.8

= 0.2

Therefore, the value of the probability that neither A nor B happen is 0.2.

 

(6) Two dice are rolled once. Find the probability of getting an even number on the first die or a total of face sum 8.

Solution:

Sample space = {(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)}

n(S) = 36

Let, A1 be the event of getting an even number on the first die

A1 = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

n (A1) = 18

∴P(A1) = n(A1)/n(S) = 18/36 = 1/2 —– (i)

Let, A2 be the event of getting a total of face sum 8.

∴ A2 = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}

n(A2) = 5

P (A2) = n(A2)/n(S) = 5/36 —- (ii)

Now, A1∩ A2 = {(2, 6), (4, 4) (6, 2)}

n (A1 ∩ A2) = 3

P (A1 ∩ A2) = n(A1∩A2)/n(S) = 3/36 = 1/12

∴P (A1∩A2) = 1/12 —– (iii)

Now, P (A1∪ A2) = P (A1) + P(A2) – P (A1∩A2)

= 1/2 + 5/36 – 1/12 [By equation (i), (ii), (iii)]

= 18+5-3/36

= 20/36

= 5/9

Therefore, the probability of getting an even number on the first die or a total of face sum 8 is 5/9.

 

(7) From a well-shuffled pack of 52 cards, a card is drawn at random. Find the probability of it being either a red king or a black queen.

Solution:

Given that n(S) = 52

We know that red king card number = 2

Black queen card number = 2

Let, A1 be the event of getting a red king card

∴n (A1) = 2

P (A1) = n(A1)/n(S) = 2/52 = 1/26

Let A2 be the event of getting a black queen card.

∴n (A2) = 2

P (A2) = n(A2)/n(S) = 2/52 = 1/26

Now, A1 and A2 are two mutually exclusive events.

∴ P (A1∩ A2) = ⌀

Now, P (A1∪ A2)

= P (A1) + P (A2) – P (A1∩A2)

= 1/26 + 1/26 – 0

= 2/26

= 1/13

Therefore, the probability of it being either a red king or a black queen is 1/13

 

(8) A box contains cards numbered 3, 5, 7, 9, … 35, 37. A card is drawn at random from the box. Find the probability that the drawn card have either multiples of 7 or a prime number.

Solution:

Sample space = {3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37}

n (s) = 18

Let, A1 be the events of getting multiples of 7

A1 = {7, 21, 35}

n (A1) = 3

P (A1) = n(A1)/n(s) = 3/18 = 1/6

Let, A2 be the events of getting a prime number.

A2 = {3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37}

n (A2) = 11

P (A2) = n(A2)/n(S) = 11/18

Now, A1∩A2 = {7}

n (A1 ∩ A2} = 1

P (A1 ∩ A2) = n(A1∩A2)/n(s) = 1/18

Now, P (A1∪ A2)

= P (A1) + P (A2) – P (A1∩A2)

= 1/6 + 11/18 – 1/18

= 3+11-1/18

= 13/18

Therefore, the probability that the drawn card have either multiples of 7 or a prime number is 13/18.

 

(9) Three unbiased coins are tossed once. Find the probability of getting atmost 2 tails or atleast 2 heads.

Solution:

Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

n (S) = 8

Let, A1 be the event of getting at most two tails.

A1 = {HHH, HHT, HTH, HTT, THH, THT, TTH}

n (A1) = 7

P (A1) = n(A1)/n(S) = 7/8

∴ P (A1) = 7/8 —– (i)

Let, A2 be the event of getting atmost two heads

∴ A2 = {HHH, HHT, HTH, THH}

∴n (A2) = 4

P (A2) = n(A1)/n(S) = 4/8 = 1/2

∴ P (A2) = 1/2 —– (ii)

Now, A1∩A2 = {HHH, HHT, HTH, THH}

n (A1∩A2) = 4

∴ P (A1∩A2) = n(A1∩A1)/n(S) = 4/8 = 1/2

∴ P (A1∩A2) = 1/2 —— (iii)

Now, P (A1∪A2)

= P (A1) + P(A2) – P (A1∩A2)

= 7/8 + 1/2 – 1/2 [By equation (i), (ii) and (iii)]

= 7/8

Thus, the probability of getting atmost two tails or atleast two heads is 7/8.

 

(10) The probability that a person will get an electrification contract is 3/5 and the probability that he will not get plumbing contract is 5/8. The probability of getting at least one contract is 5/7. What is the probability that he will get both?

Solution:

Now, Let A1be the event of getting on electrification contract.

Let, A2 be the event of getting plumbing contract.

∴ Now, Given that

P (A1) = 3/5 and P (A2) = 5/8

1 – P (A2) = 5/8

P (A2) = 1 – 5/8

P (A1) = 3/8

Now, P (A1∪A2) = 5/7

We know that P (A1∪A2) = P (A1) + P(A2) – P(A1∩A2)

5/7 = 3/5 + 3/8 – P (A1∩A2)

5/7 = 24+15 /40 – P (A1∩A2)

P (A1∩A2) = 39/40 – 5/7

P (A1∩A2) = 273-200/280

P (A1∩A2) = 73/280

Therefore, the probability of get both is 73/280

 

(11) In a town of 8000 people, 1300 are over 50 years and 3000 are females. It is known that 30% of the females are over 50 years. What is the probability that a chosen individual from the town is either a female or over 50 years?

Solution:

Let, the number of females event be A1 and the number of people over 50 years be A2.

Now, total people = 8000

∴n(S) = 8000

Number of females = 3000 and number of people over 50 years = 1300

∴n (A1) = 3000

n (A2) = 1300

∴ P (A1) = n(A1)/n(S) = 3000/8000 = 3/8 —– (i)

P (A2) = n(A2)/n(S) = 1300/8000 = 13/80 —– (ii)

Now, 30% females over 50 years.

∴ 30/100 × 3000

= 900

∴n (A1∩A2) = 900

∴ P (A1∩A2) = n(A1∩A2/n(s))

∴P (A1∩A2) = 900/8000 = 9/80 —- (iii)

Now, P (A1∪A2)

= P (A1) + P(A2) – P (A1∩A1)

= 3/8 + 13/80 – 9/80 [By equation (i), (ii) and (iii)]

= 30+13-9/80

= 34/80

= 17/40

Therefore, the probability of getting either a female or over 50 years is 17/40.

 

(12) A coin is tossed thrice. Find the probability of getting exactly two heads or atleast one tail or two consecutive heads.

Solution:

Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

n (S) = 8

Let, the getting exactly two heads event be A1

A1 = {HHT, HTH, THH}

n (A1) = 3

∴ P (A1) = n(A1)/n(S) = 3/8 —— (i)

Let, the getting at least one tail event be A2

A2 = {HHT, HTH, HTT, THH, THT, TTH, TTT}

n (A2) = 7

∴P (A2) = n(A1)/n(S) = 7/8 —- (ii)

Let, the getting two consecutive heads events be A3

∴ A3 = {HHH, HHT, THH}

n (A3) = 3

P (A3) = n(A3)/n(S) = 3/8 —- (iii)

Now, (A1∩A2) = {HHT, THH}

n (A1∩A2) = 3

∴ P (A1∩A2) = n(A1∩A2)/n(S) = 3/8 —– (iv)

Now, A2∩A3 = {HHT, THH}

n (A2∩A3) = 2

∴ P (A2∩A3) = n(A2∩A3)/n(S) = 2/8 —-  (v)

Now, A3∩A1 = {HHT, THH}

n (A3∩A1) = 2

P (A3∩A2) = n(A3∩A1)/n(s) = 2/8 —- (vi)

Now, A1∩A2∩A3 = {HHT, THH}

n (A1∩A2A3) = 2

P (A1∩A2∩A3) = n(A1∩A2∩A3)/n(S) = 2/8 —- (vii)

Now, P (A1∪A2∪A3)

= P (A1) + P (A2) + P (A3) + P (A1∩A2∩A3) – P (A1∩A2) – P (A2∩A3) – P (A3∩A1)

= 3/8 + 7/8 + 3/8 + 2/8 – 3/8 – 2/8 – 2/8 [By equation (i), (ii), (iii), (iv), (v), (vi) and (vii)]

= 3+7-2/8

= 8/8

= 1

Therefore, the probability of getting exactly two heads or atleast one tail or two consecutive heads is 1.

 

(13) If A, B, C are any three events such that probability of B is twice as that of probability of A and probability of C is thrice as that of probability of A and if P(A∩B) = 1/6, P(B∩C) = 1/4, P(A∩C) = 1/8, P(A∪B∪C) = 9/10, P(A∩B∩C) = 1/15, then find P(A), P(B) and P(C)?

Solution:

Let, P(A) = x

P(B) = 2x

P(C) = 3x

Given that, P(A∩B) – 1/6, P (B∩C) = 1/4, P (A∩C) = 1/8, P (A∪B∪C) = 9/10, P (A∩B∩C) = 1/15

We know that,

P (A∪B∪C) = p(A) + P(B) + P(C) + P(A∩B∩C) – P(A∩B) – P (B∩C) – P (A∩C)

9/10 = x + 2x + 3x + 1/15 – 1/6 – 1/4 – 1/8

9/10 = 6x + 1/15 – 1/6 – 1/4 – 1/8

6x = 9/10 – 1/15 + 1/6 + 1/4 + 1/8

6x = 108-8+20+30+15/120

x = 165/120 x 6

x = 11/48

Thus, P (A) = 11/48, P(B) = 2×11/48 = 11/24 and P(C) = 3 × 11/48 = 11/16

 

(14) In a class of 35, students are numbered from 1 to 35. The ratio of boys to girls is 4:3. The roll numbers of students begin with boys and end with girls. Find the probability that a student selected is either a boy with prime roll number or a girl with composite roll number or an even roll number.

Solution:

Now, sample space = {1, 2, 3, — 35}

∴n(S) = 35

Now, boys and gives ratio 4:3

∴ Boys = 4/7 × 35 = 20

Girls = 3/7 × 35 = 15

Boys roll number = {1, 2, 3 —– 20}

Gives roll number = {21, 22, —— 35}

Let, getting a boy roll number with prime number event be A1

∴ A1 = {2, 3, 5, 7, 11, 13, 7, 19}

n (A1) = 8

P (A1) = n(A1)/n(S) = 8/35 —- (i)

Let getting girls roll number with composite number event be A2

A2 = {21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35}

n (A2) = 12

∴ P (A2) = n (A1)/n (S) = 12/35

∴P (A2) = 12/35 event (ii)

Let, getting an even roll number event be A3

A3 = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34}

n (A3) = 17

∴ P (A3) = n(A3)/n(S) = 17/35 —- (iii)

Now, A1∩A2 = {0}

n (A1∩A2) = 0

P (A1∩A2) = ⌀ —– (iv)

Now, A2∩A3 = {22, 24, 26, 28, 30, 32, 34}

n (A2∩A3) = 7

∴ P (A2∩A3) = n(A2∩A2)/n(s) = 7/35 —– (v)

Now, A1∩A3 = {2 –}

n (A1∩A3) = 1

∴ P (A1∩A3) = n(A1∩A3)/n(3) = 1/35 —- (vi)

Now, A1∩A2∩A3 = {0}

n (A1∩A2∩A3) = 0 ∴ P (A1∩A2∩A3) = ⌀ —- (vii)

Now, P (A1∪A2∪A3)

= P (A1) + P(A2) + P(A3) + P(A1∩A2∩A3) – P(A1∩A2) – P(A2∩A3) – P(A1∩A3)

= 8/35 + 12/35 + 17/35 + 0 – 0 – 0 – 7/35 – 1/35 [By equation (i), (ii), (iii), (iv), (v), (vi) or (vii)]

= 8+12+17-7-1/35 = 29/35

∴ P (A1∪A2∪A3) = 29/35

Therefore, the probability of getting either a boy, with prime roll number or a girl with composite roll number or an even roll number is 29/35.

 

Here is your solution of TN Samacheer Kalvi Class 10 Math Chapter 8 Statistics and Probability Exercise 8.4

Chapter 1 Relations and Functions Chapter 2 Numbers and Sequences
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Exercise 4.2 Solution

Exercise 4.3 Solution

Exercise 4.4 Solution

Exercise 4.5 Solution

Exercise 4 Solution

 

Chapter 5 Coordinate Geometry Chapter 6 Trigonometry
Exercise 5.1 Solution

Exercise 5.2 Solution

Exercise 5.3 Solution

Exercise 5.4 Solution

Exercise 5.5 Solution

Unit Exercise 5 Solution

 

Exercise 6.1 Solution

Exercise 6.2 Solution

Exercise 6.3 Solution

Exercise 6.4 Solution

Exercise 6.5 Solution

Unit Exercise 6 Solution

Chapter 7 Mensuration Chapter 8 Statistics and Probability
Exercise 7.1 Solution

Exercise 7.2 Solution

Exercise 7.3 Solution

Exercise 7.4 Solution

Exercise 7.5 Solution

Unit Exercise 7 Solution

Exercise 8.1 Solution

Exercise 8.2 Solution

Exercise 8.3 Solution

Exercise 8.4 Solution

Exercise 8.5 Solution

Unit Exercise 8 Solution

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Updated: March 4, 2022 — 2:43 pm

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