# 10th Maths Book Back Statistics and Probability Ex 8.4

## Samacheer Kalvi 10th Maths Book Back Solution:

Tamil Nadu 10th Maths Book Back Answers Unit 8 – Statistics and Probability Ex 8.4 are provided on this page. Samacheer Kalvi Maths Book Back Solutions/ Guide available for all Units. TN Samacheer Kalvi 10th Maths Book consists of 8 Units and each unit book back solutions given below topics wise with Questions and Answers. The complete Samacheer Kalvi Books Back Answers/Solutions are available on our site.

The Samacheer kalvi 10th Maths solutions are useful to enhance your skills. Candidates who prepared for the Competitive and board exams 10th Maths Book Back Answers in English and Tamil Medium. The 10th Maths Unit 8 Statistics and Probability consist of 5 units. Each Unit Book Back Answers provide topic-wise on this page. 10th Maths Book Back Answers are prepared according to the latest syllabus. The 10th Maths Book Back Relations and Functions Ex 8.4 Answers in English.

### 10th Maths Book Back Answers/Solutions:

TN Samacheer Kalvi 10th Maths Chapter 8 Book Back Exercise given below. The 10th Maths Book Back Solutions Guide is uploaded below:

### Exercise 8.4 Statistics and Probability

1.If P(A) = $$\frac{2}{3}$$, P(B) = $$\frac{2}{5}$$, P(A∪B) = 13 then find P(A∩B).
Solution:
P(A) = $$\frac{2}{3}$$, P(B) = $$\frac{2}{5}$$, P(A∪B) = $$\frac{1}{3}$$
P(A ∩ B) = P(A) + P(B) – P(A∪B)

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)
(it) P(not B)
(iii) P(A or B)
(i) P(not A) = 1 – P (A) = 1 – 0.42 = 0.58
(ii) P(not B) = 1 – P(B) = 1 – 0.48 = 0.52
(iii) P(A or B) = P(A) + P(B) – P(A ∩ B) = 0.42 + 0.48 – 0.16 = 0.90 – 0.16 = 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:
A and B are two mutually exclusive events of a random experiment.
P(not A) = 0.45,
P(A) = 1 – P(not A)
P(A∪B) = 0.65 = 1 – 0.45 = 0.55
P(A∪B) = P(A) + P(B) = 0.65
0.55 + P(B) = 0.65
P(B) = 0.65 – 0.55
= 0.10

4.The probability that atleast one of A and B occur is 0.6. If A and B occur simultaneously with probability 0.2, then find P($$\bar{A}$$) + P($$\bar{B}$$).
Here P (A ∪ B) = 0.6, P (A ∩ B) = 0.2
P (A ∪ B) = P (A) + P (B) – P (A ∩ B)
0.6 = P (A) + P (B) – 0.2
P(A) + P(B) = 0.8
P($$\bar{A}$$) + P($$\bar{B}$$) = 1 – P(A) + 1 – P(B)
= 2 – [P(A) + P(B)]
= 2 – 0.8
= 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:
P(A) = 0.5 Since A and B are mutually inclusive events
P(B) = 0.3 events.
P($$\overline{\mathbf{A}}$$)∪P($$\overline{\mathbf{B}}$$) = 1 – [P(A) + P(B)]
= 1 – [0.5 + 0.3] = 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:
Two dice rolled once.

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:
n(S) = 52
No. of Red cards = 26,
Red king cards = 2
No. of Black cards = 26,
Black queen cards = 2
No. of red king cards = n(K) = 2

∴ The probability of being either a red king or a black queen = $$\frac{1}{13}$$.

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 has either multiples of 7 or a prime number.
Solution:
S = {3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37}
n(S) = 18
Multiplies of seven cards (A) = {7, 21, 35}
= n(A) = 3

Let the prime number cards B
B = {3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37}
n(B) = 11

9.Three unbiased coins are tossed once. Find the probability of getting at most 2 tails or at least 2 heads.
Solution:
When we toss three coins, the sample space S = {HHH, TTT, HTT, THH, HHT, TTH, HTH, THT}
n(S) = 8
Event of getting at most 2 tails be A.
∴ A = { HHH, HTT, THH, HHT, TTH, HTH, THT}

10. The probability that a person will get an electrification contract is a probability that he will not get a plumbing contract is $$\frac{3}{5}$$. The probability of getting at least one contract is $$\frac{5}{8}$$. What is the probability that he will get both?
Solution:

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:

12. A coin is tossed thrice. Find the probability of getting exactly two heads or at least one tail or two consecutive heads.
Solution:
Three coins tossed simultaneously.
S = { HHH, TTT, HHT, TTH, HTH, THT, HTT, THH}
n(S) = 8
Happening of getting exactly two heads be A.
A= {HHT, HTH, THH}
n(A) = 3

Event of getting at least one tail be B.
∴ B = {TTT, HHT, TTH, HTH, THT, HTT, THH}

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) = $$\frac{1}{6}$$, P(B∩C) = $$\frac{1}{4}$$, P(A∩C) = $$\frac{1}{8}$$, $$\mathbf{P}(\mathbf{A} \cup \mathbf{B} \cup \mathbf{C})=\frac{9}{10}, \mathbf{P}(\mathbf{A} \cap \mathbf{B} \cap \mathbf{C})=\frac{1}{15}$$, then find P(A), P(B) and P(C)?
Solution:
P(B) = 2P(A)
P(C) = 3P(A)

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 a prime roll number or a girl with a composite roll number or an even roll number.
Solution: