RS Aggarwal Class 8 Mathematics Solutions Chapter 24 Probability

Probability

1. Experiment:

An action that can lead to clearly defined results is referred to as an experiment.

2. Random Experiment:

An action in which all probable results are recognized and the specific result cannot be determined beforehand is referred to as a random experiment.

Examples:

• Rolling an unbiased die.
• Tossing a fair coin.
• Drawing a card from a well-shuffled pack.
• Picking up a ball of a particular colour from a bag holding balls of various colours.

Details:

• When we flip a coin, either a Head (H) or a Tail (T) will show up.
• A die is a solid cube with 6 faces, marked 1, 2, 3, 4, 5, 6 respectively. When we roll a die, the result shown is the number on its top surface.
• A pack of cards contains 52 cards. It includes 13 cards in each suit: Spades, Clubs, Hearts, and Diamonds. Spades and clubs are black cards. Hearts and diamonds are red cards. There are 4 copies of each unit. Kings, Queens, and Jacks are present. These are all referred to as face cards.

3. Sample Space:

When we carry out an action, the group of all probable results is called the sample space.

Examples:

• In tossing a coin, S = {H, T}
• If two coins are tossed, the S = {HH, HT, TH, TT}
• In rolling a die, we have, S = {1, 2, 3, 4, 5, 6}

4. Event:

Any subset of a sample space is referred to as an event.

5. Probability of Occurrence of an Event:

Let S be the sample and let E be an event.
Then, E ⊂ S

\[ \therefore P(E) = \frac{n(E)}{n(S)} \]

6. Results on Probability:

• P(S) = 1
• 0 ≤ P(E) ≤ 1
• P(∅) = 0
• For any events A and B we have: P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
• If A denotes (not-A), then P(A) = 1 - P(A)

Question 1. When we toss one coin, what are all the possible outcomes? When we toss two coins, list all the possible outcomes. When we roll a die, what are the possible outcomes? What is the total number of possible outcomes when we pick a card from a 52-card deck?
Answer: (i) When we flip one coin, the outcomes that are possible are head (H) and tail (T).
(ii) When we flip two coins, the possible results are HH, HT, TH and TT.
(iii) When we roll a die, the outcomes possible are 1, 2, 3, 4, 5 and 6.
(iv) The total count of possible outcomes is 52.

Exam Tip: Always list all outcomes systematically - don't skip any. For coin tosses, use the table method to ensure you capture every combination.

Question 2. A coin is flipped once. Find the probability that a tail appears.
Answer: The outcomes that can occur from a coin flip are H and T.
Total count of outcomes = 2
Number of tails = 1
\[ \therefore P(\text{tail}) = \frac{1}{2} \]

Exam Tip: Always identify the total number of possible outcomes first, then count favorable outcomes - this ensures your fraction is correct.

Question 3. Two coins are tossed. Find: (i) the probability of getting both tails; (ii) the probability of getting at least one tail; (iii) the probability of getting at most one tail.
Answer: The outcomes from tossing two coins are HH, HT, TH and TT.
That is, the count of possible outcomes = 4

(i) Getting both tails means TT
Count of outcomes with two tails = 1
\[ \therefore P(\text{both tails}) = \frac{1}{4} \]

(ii) Getting at least one tail means HT, TH and TT
With at least one tail, the count of outcomes = 3
\[ \therefore P(\text{at least 1 tail}) = \frac{3}{4} \]

(iii) Getting at most one tail means HH, HT and TH
The count of outcomes with at most 1 tail = 3
\[ \therefore P(\text{at most 1 tail}) = \frac{3}{4} \]

Exam Tip: Pay close attention to the phrasing - "at least" means that value or more, while "at most" means that value or less. List all matching outcomes carefully.

Question 4. A bag has 4 white balls and 5 blue balls. If we pick one ball randomly, find: (i) the probability that it is white; (ii) the probability that it is blue.
Answer: The total count of balls = 4 + 5 = 9
(i) The count of white balls = 4
\[ \therefore P(\text{white ball}) = \frac{4}{9} \]

The count of blue balls = 5
\[ \therefore P(\text{blue ball}) = \frac{5}{9} \]

Exam Tip: Always add all items in the bag first to get the denominator. Then count the specific type you're looking for - this is your numerator.

Question 5. A bag contains 5 green balls, 6 white balls, and 4 red balls. A ball is picked at random. Find: (i) the probability of selecting a green ball; (ii) the probability of selecting a white ball; (iii) the probability of selecting a ball that is not red.
Answer: The total count of balls = 5 + 6 + 4 = 15
(i) The count of green balls = 4
\[ \therefore P(\text{green ball}) = \frac{4}{15} \]

(ii) The count of white balls = 5
\[ \therefore P(\text{white ball}) = \frac{5}{15} = \frac{1}{3} \]

(iii) The count of balls that are not red (i.e., 5 white and 4 green) = 9
\[ \therefore P(\text{non red balls}) = \frac{9}{15} = \frac{3}{5} \]

Exam Tip: For "not red," add all the other colors together rather than subtracting. This prevents calculation errors and is clearer to examiners.

Question 6. In a raffle draw, there are 10 winning tickets and 20 non-winning tickets. If one ticket is drawn at random, find the probability of getting a prize.
Answer: The total count of tickets = 10 + 20 = 30
The count of prize tickets = 10
\[ \therefore P(\text{getting a prize}) = \frac{10}{30} = \frac{1}{3} \]

Exam Tip: Always simplify your fraction to its lowest form - examiners expect the final answer in simplest terms.

Question 7. A box contains 100 light bulbs. Among them, 8 bulbs are broken. Find: (i) the probability that a randomly selected bulb is defective; (ii) the probability that a randomly selected bulb is working.
Answer: The total count of bulbs in the box = 100
(i) The count of defective bulbs = 8
\[ \therefore P(\text{defective bulb}) = \frac{8}{100} = \frac{2}{25} \]

(ii) The count of functioning bulbs = 100 - 8 = 92
\[ \therefore P(\text{non defective bulb}) = \frac{92}{100} = \frac{23}{25} \]

Exam Tip: Always verify: P(defective) + P(working) should equal 1. This quick check catches arithmetic mistakes before you submit.

Question 8. A spinner wheel has sectors. If the wheel has 3 red sectors and 5 green sectors, find the probability of spinning a green sector. (b) The wheel contains 5 + 3 = 8 sectors total. The count of green sectors = 5. Now, the probability of landing on a green sector = 5/8.
Answer: The spinner has a total of 5 + 3 = 8 sectors. The count of green sectors = 5. Therefore, \[ P(\text{getting a green sector}) = \frac{5}{8} \]

Exam Tip: For spinner problems, count all the sections carefully - missed sections lead to incorrect denominators.

Question 9. A survey was carried out among 200 women. It was discovered that 118 women do not like coffee. If a woman is chosen at random, what is the probability that she dislikes coffee?
Answer: The total count of ladies surveyed = 200
The count of ladies who dislike coffee = 118

If a lady is picked randomly, \[ P(\text{a lady that dislikes coffee}) = \frac{118}{200} = \frac{59}{100} \]

Exam Tip: Always simplify large fractions by finding the GCD - this shows mathematical maturity and gives the cleaner final form examiners prefer.

Question 10. A number is selected randomly from the numbers between 1 and 19. Find: (i) the probability that the selected number is prime; (ii) the probability that the selected number is even; (iii) the probability that the selected number is divisible by 3.
Answer: The total count of possible outcomes = 19
(i) The prime numbers from 1 to 19 are 2, 3, 5, 7, 11, 13, 17 and 19.
The total count of primes = 8
\[ \therefore P(\text{prime number}) = \frac{8}{19} \]

(ii) The even numbers from 1 to 19 are 2, 4, 6, 8, 10, 12, 14, 16 and 18.
The total count of even numbers = 6
\[ \therefore P(\text{even number}) = \frac{6}{19} \]

(iii) The numbers from 1 to 19 that are divisible by 3 are 3, 6, 9, 12, 15 and 18.
The total count of possible outcomes = 6
\[ \therefore P(\text{number divisible by 3}) = \frac{6}{19} \]

Exam Tip: For divisibility questions, list the multiples systematically - use the multiplication table method to avoid missing any.

Question 11. A card is drawn randomly from a standard pack of 52 cards. Find: (i) the probability of drawing a king; (ii) the probability of drawing a spade; (iii) the probability of drawing a red queen; (iv) the probability of drawing a black 8.
Answer: The total count of possible outcomes = 52
(i) There are 4 king cards (king of hearts, king of diamonds, king of spades and king of cloves)
The count of kings = 4
\[ \therefore P(\text{king}) = \frac{4}{52} = \frac{1}{13} \]

(ii) There is a group of 13 spade cards.
The count of spades = 13
\[ \therefore P(\text{spades}) = \frac{13}{52} = \frac{1}{4} \]

(iii) There are 2 red queens in a pack (queen of hearts and queen of diamonds)
The count of red queens = 2
\[ \therefore P(\text{red queen}) = \frac{2}{52} = \frac{1}{26} \]

(iv) There are 2 black 8s in a pack (8 of cloves and 8 of spades)
The count of black 8s = 2
\[ \therefore P(\text{black 8}) = \frac{2}{52} = \frac{1}{26} \]

Exam Tip: Memorize the standard deck structure - 4 suits of 13 cards each, with specific face cards. This speeds up card probability problems significantly.

Question 12. A standard deck of 52 cards is drawn. Find: (i) the probability of drawing a card with the number 4; (ii) the probability of drawing a queen; (iii) the probability of drawing a black card.
Answer: The total count of possible outcomes = 52
(i) There are 4 cards with the number 4 (4 of hearts, 4 of diamonds, 4 of spades and 4 of cloves)
\[ \therefore P(\text{4 card}) = \frac{4}{52} = \frac{1}{13} \]

(ii) There are 4 queens in a pack of cards (queen of hearts, queen of diamonds, queen of spades and queen of cloves)
\[ \therefore P(\text{queen}) = \frac{4}{52} = \frac{1}{13} \]

(iii) There is a combined total of 26 black cards (13 spade cards and 13 clove cards)
\[ \therefore P(\text{black card}) = \frac{26}{52} = \frac{1}{2} \]

Exam Tip: For deck problems, organize your counting - separate the face cards from number cards and the suits by color to avoid double-counting.

Question 1. A spinner wheel is split into a certain number of sectors. If the wheel is separated into 5 + 3 = 8 sectors total, with 5 being green sectors, determine the probability of the pointer landing on a green sector.
Answer: (b) \( \frac{5}{8} \)
The spinner has a total of 5 + 3 = 8 sectors.
The count of green sectors = 5
Therefore, \[ P(\text{getting a green sector}) = \frac{5}{8} \]

Exam Tip: Always count the total sectors first - this prevents errors in setting up your probability fraction.

Question 2. A set of cards numbered 1 through 8 is available. What is the probability of picking a card with a number less than 4?
Answer: (c) \( \frac{3}{8} \)
The total count of cards = 8
Cards with numbers smaller than 4 = 3 (cards numbered 1, 2 and 3)
Therefore, \[ P(\text{getting a number less than 4}) = \frac{3}{8} \]

Exam Tip: For range-based questions, be precise about whether the boundary is included ("less than" vs. "less than or equal to").

Question 3. Two coins are flipped. What is the probability that the outcome includes one head and one tail?
Answer: (b) \( \frac{1}{2} \)
When we flip two coins, the outcomes possible are HH, HT, TH and TT.
The total count of outcomes = 4
The count of outcomes containing one head and one tail = 2
Therefore, \[ P(\text{one head and one tail}) = \frac{2}{4} = \frac{1}{2} \]

Exam Tip: Order matters in coin tosses - HT and TH are different outcomes. List them separately to get the correct count.

Question 4. A bag holds balls of various colors - some red and some of other colors - for a total of 5. If there are 2 red balls, what is the probability of drawing a red ball?
Answer: (d) \( \frac{2}{5} \)
The total count of outcomes = 5
The count of red balls = 2
Therefore, \[ P(\text{red ball}) = \frac{2}{5} \]

Exam Tip: Always work with the numbers given - don't make assumptions about missing information. Use only what the problem states.

Question 5. A die is rolled. What is the probability of rolling a 6?
Answer: (b) \( \frac{1}{6} \)
The outcomes when rolling a die are 1, 2, 3, 4, 5 and 6.
The total count of outcomes = 6
Therefore, \[ P(\text{getting 6}) = \frac{1}{6} \]

Exam Tip: For single-event dice problems, the denominator is always 6 - the number of sides. Focus on counting favorable outcomes correctly.

Question 6. A die is rolled. What is the probability of rolling an even number?
Answer: (a) \( \frac{1}{2} \)
The total count of outcomes = 6 (Numbers: 1, 2, 3, 4, 5 and 6)
The even numbers are 2, 4, and 6.
The count of favorable outcomes = 3
Therefore, \[ P(\text{even number}) = \frac{3}{6} = \frac{1}{2} \]

Exam Tip: Group outcomes by the property you're looking for (even, odd, divisible by, etc.) before counting - this reduces errors.

Question 7. A die is rolled. Find the probability of rolling an odd number.
Answer: The total count of possible outcomes when rolling a die is 6. The odd numbers among 1 to 6 are 1, 3, and 5. The count of odd outcomes = 3. Therefore, the probability of rolling an odd number = \( \frac{3}{6} = \frac{1}{2} \).

Exam Tip: Odd and even outcomes on a standard die are always equal in count - each has probability 1/2. Use this as a quick verification.

Question 8. A card is drawn from a standard 52-card deck. What is the probability of drawing a queen?
Answer: (c) 113
The total count of cards = 52
The count of queens = 4 (i.e., queen of hearts, queen of diamonds, queen of cloves and queen of spades)
Therefore, \[ P(\text{queen}) = 452 = 113 \]

Exam Tip: Each rank (king, queen, jack, ace, number cards) appears exactly 4 times in a standard deck - one per suit. Memorize this fact.

Question 9. A card is selected from a standard pack of 52 cards. What is the probability of drawing a black 6?
Answer: (b) \( \frac{1}{26} \)
The total count of cards = 52
The count of black 6 cards = 2 (6 of spades, 6 of cloves)
Therefore, \[ P(\text{black 6}) = \frac{2}{52} = \frac{1}{26} \]

Exam Tip: Black cards are spades and clubs only - remember this distinction as it's often tested in card probability problems.