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## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 8 Statistics and Probability Ex 8.4

Question 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)

Question 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)

Answer:

(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

Question 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

Question 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(\(\bar{B}\)).

Answer:

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

Question 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

Question 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.

Question 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 = \(\frac{1}{13}\).

Question 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

Question 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}

Question 10.

The probability that a person will get an electrification contract is the probability that he will not get 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:

Question 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:

Question 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}

Question 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)

Question 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:

n(S) = 35