Math Notes For Class 8 Chapter 6 Ex 6.1 Solved & Quiz

Question 1 (Part i): A fair coin is tossed once. Find the probability of getting a Head.

Solution:
When a single coin is tossed, the sample space $S$ is:
$$S = \{\text{Head}, \text{Tail}\} = \{H, T\}$$
Total number of possible outcomes, $n(S) = 2$.
Let $A$ be the event of getting a Head:
$$A = \{H\}$$
Number of favorable outcomes, $n(A) = 1$.
Using the formula for probability:
$$P(A) = \frac{n(A)}{n(S)} = \frac{1}{2}$$
Answer: $$\frac{1}{2}$$

Question 1 (Part ii): Two fair coins are tossed simultaneously. Find the probability of getting at least one Head.

Solution:
When two coins are tossed together, the sample space $S$ is:
$$S = \{(H,H), (H,T), (T,H), (T,T)\}$$
Total number of outcomes, $n(S) = 4$.
Let $B$ be the event of getting at least one Head (1 or 2 Heads):
$$B = \{(H,H), (H,T), (T,H)\}$$
Number of favorable outcomes, $n(B) = 3$.
Calculating probability:
$$P(B) = \frac{n(B)}{n(S)} = \frac{3}{4}$$
Answer: $$\frac{3}{4}$$

Question 1 (Part iii): Two fair coins are tossed simultaneously. Find the probability of getting two Tails.

Solution:
From the sample space of two tossed coins:
$$S = \{(H,H), (H,T), (T,H), (T,T)\}, \quad n(S) = 4$$
Let $C$ be the event of getting two Tails:
$$C = \{(T,T)\}$$
Number of favorable outcomes, $n(C) = 1$.
$$P(C) = \frac{n(C)}{n(S)} = \frac{1}{4}$$
Answer: $$\frac{1}{4}$$

Question 2 (Part i): A six-sided fair die is rolled once. Find the probability of getting an even number.

Solution:
The sample space for rolling a die is:
$$S = \{1, 2, 3, 4, 5, 6\}$$
Total possible outcomes, $n(S) = 6$.
Let $E_1$ be the event of rolling an even number:
$$E_1 = \{2, 4, 6\}$$
Number of favorable outcomes, $n(E_1) = 3$.
$$P(E_1) = \frac{n(E_1)}{n(S)} = \frac{3}{6} = \frac{1}{2}$$
Answer: $$\frac{1}{2}$$

Question 2 (Part ii): A six-sided fair die is rolled once. Find the probability of getting a number greater than 4.

Solution:
Sample space, $S = \{1, 2, 3, 4, 5, 6\}$, $n(S) = 6$.
Let $E_2$ be the event of getting a number greater than 4:
$$E_2 = \{5, 6\}$$
Number of favorable outcomes, $n(E_2) = 2$.
$$P(E_2) = \frac{n(E_2)}{n(S)} = \frac{2}{6} = \frac{1}{3}$$
Answer: $$\frac{1}{3}$$

Question 2 (Part iii): A six-sided fair die is rolled once. Find the probability of getting a prime number.

Solution:
Sample space, $S = \{1, 2, 3, 4, 5, 6\}$, $n(S) = 6$.
Prime numbers in the sample space are $2, 3,$ and $5$.
Let $E_3$ be the event of getting a prime number:
$$E_3 = \{2, 3, 5\}$$
Number of favorable outcomes, $n(E_3) = 3$.
$$P(E_3) = \frac{n(E_3)}{n(S)} = \frac{3}{6} = \frac{1}{2}$$
Answer: $$\frac{1}{2}$$

Question 3 (Part i): A bag contains 5 red marbles, 4 green marbles, and 3 blue marbles. A marble is drawn at random. Find the probability of drawing a red marble.

Solution:
Total number of marbles in the bag:
$$n(S) = 5 + 4 + 3 = 12$$
Let $R$ be the event of drawing a red marble.
Number of red marbles, $n(R) = 5$.
$$P(R) = \frac{n(R)}{n(S)} = \frac{5}{12}$$
Answer: $$\frac{5}{12}$$

Question 3 (Part ii): Using the same bag (5 red, 4 green, 3 blue), find the probability of drawing either a green or a blue marble.

Solution:
Total outcomes, $n(S) = 12$.
Let $G \cup B$ be the event of drawing a green or blue marble.
Number of green or blue marbles $= 4 + 3 = 7$.
Therefore, $n(G \cup B) = 7$.
$$P(G \cup B) = \frac{7}{12}$$
Answer: $$\frac{7}{12}$$

Question 3 (Part iii): Using the same bag (5 red, 4 green, 3 blue), find the probability of NOT drawing a red marble.

Solution:
From Part (i), $P(R) = \frac{5}{12}$.
Using the complement probability formula $P(R') = 1 - P(R)$:
$$P(R') = 1 - \frac{5}{12} = \frac{12 - 5}{12} = \frac{7}{12}$$
Alternatively, non-red marbles $= 4 \text{ green} + 3 \text{ blue} = 7$.
$$P(R') = \frac{7}{12}$$
Answer: $$\frac{7}{12}$$

Question 4 (Part i): A card is drawn at random from a standard deck of 52 playing cards. Calculate the probability of drawing an Ace.

Solution:
Total number of cards in a deck, $n(S) = 52$.
Total number of Aces in a standard deck $= 4$.
Let $A$ be the event of drawing an Ace, so $n(A) = 4$.
$$P(A) = \frac{n(A)}{n(S)} = \frac{4}{52} = \frac{1}{13}$$
Answer: $$\frac{1}{13}$$

Question 4 (Part ii): A card is drawn from a 52-card deck. Calculate the probability of drawing a black card.

Solution:
Total cards, $n(S) = 52$.
A standard deck contains 26 red cards and 26 black cards (Spades and Clubs).
Let $B$ be the event of drawing a black card, so $n(B) = 26$.
$$P(B) = \frac{n(B)}{n(S)} = \frac{26}{52} = \frac{1}{2}$$
Answer: $$\frac{1}{2}$$

Question 4 (Part iii