RS Aggarwal Class 9 Mathematics Solutions Chapter 15 Probability

Probability

1. Experiment: An operation which can produce some well-defined outcomes is called an experiment.

2. Random Experiment: An experiment in which all possible outcomes are known and the exact output cannot be predicted in advance is called a random experiment.

• Rolling an unbiased die.
• Tossing a fair coin.
• Drawing a card from a pack of well-shuffled cards.
• Picking up a ball of certain colour from a bag containing balls of different colours.

• When we throw a coin, then either a Head (H) or a Tail (T) appears.
• A die is a solid cube, having 6 faces, marked 1, 2, 3, 4, 5, 6 respectively. When we throw a die, the outcome is the number that appears on its upper face.
• A pack of cards has 52 cards. It has 13 cards of each suit, name Spades, Clubs, Hearts and Diamonds. Cards of spades and clubs are black cards. Cards of hearts and diamonds are red cards. There are 4 honours of each unit. There are Kings, Queens and Jacks. These are all called face cards.

3. Sample Space: When we perform an experiment, then the set S of all possible outcomes 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 called an event.

5. Probability of Occurrence of an Event: Let S be the sample and let E be an event. Then, \( E \subseteq S \)
\( \therefore P(E) = \frac{n(E)}{n(S)} \)

6. Results on Probability:

• (i) P(S) = 1
• (ii) \( 0 \leq P(E) \leq 1 \)
• (iii) P(ϕ) = 0
• (iv) For any events A and B we have \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
• (v) If Ā denotes (not-A), then \( P(\bar{A}) = 1 - P(A) \)

Exercise 15(A)

Question 1. A coin is tossed 500 times and we get heads 285 times and tails 215 times. (i) What is the probability of getting a head? (ii) What is the probability of getting a tail?
Answer: Given: Total number of trials = 500, Number of heads = 285, Number of tails = 215
(i) The probability of getting a head is calculated by dividing the count of heads by the total count of trials: \( P(\text{head}) = \frac{285}{500} = 0.57 \)
(ii) The probability of getting a tail is calculated by dividing the count of tails by the total count of trials: \( P(\text{tail}) = \frac{215}{500} = 0.43 \)

Exam Tip: Always add the number of heads and tails to verify they sum to the total trials - this checks your work before calculating probabilities.

Question 2. Two coins are tossed randomly 400 times, and the results show 2 heads 112 times, 1 head 160 times, and 0 heads 128 times. Find the probability of: (i) getting 2 heads; (ii) getting 1 head; (iii) getting 0 heads.
Answer: Given: Total number of trials = 400, Frequency of 2 heads = 112, Frequency of 1 head = 160, Frequency of 0 heads = 128
(i) The probability of getting 2 heads: \( P(E_1) = \frac{112}{400} = 0.28 \)
(ii) The probability of getting 1 head: \( P(E_2) = \frac{160}{400} = 0.4 \)
(iii) The probability of getting 0 heads: \( P(E_3) = \frac{128}{400} = 0.32 \)

Exam Tip: Verify your answer by adding all three probabilities - they should equal 1, confirming your calculations are complete and correct.

Question 3. Three coins are tossed at random 200 times. The results show 3 heads 39 times, 2 heads 58 times, 1 head 67 times, and 0 heads 36 times. Find the probability of: (i) getting 3 heads; (ii) getting 1 head; (iii) getting 0 heads; (iv) getting 2 heads.
Answer: Given: Total number of trials = 200, Number of times 3 heads appeared = 39, Number of times 2 heads appeared = 58, Number of times 1 head appeared = 67, Number of times 0 heads appeared = 36
(i) The probability of getting 3 heads: \( P(E_1) = \frac{39}{200} = 0.195 \)
(ii) The probability of getting 1 head: \( P(E_2) = \frac{67}{200} = 0.335 \)
(iii) The probability of getting 0 heads: \( P(E_3) = \frac{36}{200} = 0.18 \)
(iv) The probability of getting 2 heads: \( P(E_4) = \frac{58}{200} = 0.29 \)

Exam Tip: Always verify that the sum of all frequencies matches the total number of trials before proceeding with probability calculations.

Question 4. A die is thrown 300 times and the frequencies of outcomes 3, 6, 5, and 1 are 54, 33, 39, and 60 times respectively. Find the probability of getting: (i) 3; (ii) 6; (iii) 5; (iv) 1.
Answer: Given: Total number of trials = 300, The events are E₁, E₂, E₃, and E₄ representing outcomes 3, 6, 5, and 1 respectively.
(i) The probability of getting 3: \( P(E_1) = \frac{54}{300} = 0.18 \)
(ii) The probability of getting 6: \( P(E_2) = \frac{33}{300} = 0.11 \)
(iii) The probability of getting 5: \( P(E_3) = \frac{39}{300} = 0.13 \)
(iv) The probability of getting 1: \( P(E_4) = \frac{60}{300} = 0.2 \)

Exam Tip: For a fair die, each outcome should have a probability close to 1/6 ≈ 0.167. Compare your calculated values to this theoretical value.

Question 5. In a survey of 200 women, 142 like coffee and 58 do not like coffee. If a lady is chosen at random, find the probability that: (i) the selected woman likes coffee; (ii) the selected woman does not like coffee.
Answer: Given: Total number of women = 200, Number of women who like coffee = 142, Number of women who do not like coffee = 58
(i) Let E₁ be the event that the chosen woman likes coffee. Then: \( P(E_1) = \frac{142}{200} = 0.71 \)
(ii) Let E₂ be the event that the chosen woman dislikes coffee. Then: \( P(E_2) = \frac{58}{200} = 0.29 \)

Exam Tip: Note that P(E₁) + P(E₂) = 1, which confirms the complementary nature of these two events - one person either likes or dislikes coffee, with no middle ground.

Question 6. A student's test scores in 6 exams show that he scores more than 60% marks in 2 tests. Find the probability that he will score more than 60% marks in the next test.
Answer: Given: Number of tests in which he gets more than 60% marks = 2, Total number of tests = 6
Required probability: \( P = \frac{2}{6} = \frac{1}{3} \)

Exam Tip: This uses empirical probability based on past performance - the relative frequency of success provides an estimate for future occurrences.

Question 7. In a parking lot, there are 240 vehicles, of which 84 are two-wheelers. Find the probability that a randomly selected vehicle is a two-wheeler.
Answer: Given: Total number of vehicles = 240, Number of two-wheelers = 84
Required probability = \( \frac{\text{Number of two-wheelers}}{\text{Total number of vehicles}} = \frac{84}{240} = 0.35 \)

Exam Tip: Always simplify fractions when possible - here 84/240 can be reduced to 7/20, making mental calculations easier.

Question 8. A telephone directory contains 200 phone numbers. Of these, 24 have a units digit of 5 and 16 have a units digit of 8. Find the probability that: (i) a randomly selected number has a units digit of 5; (ii) a randomly selected number has a units digit of 8.
Answer: Given: Total phone numbers = 200, Numbers with units digit 5 = 24, Numbers with units digit 8 = 16
(i) Required probability = \( \frac{24}{200} = 0.12 \)
(ii) Required probability = \( \frac{16}{200} = 0.08 \)

Exam Tip: Notice that 0.12 + 0.08 = 0.2, meaning 20% of numbers have either digit 5 or 8 as their units digit.

Question 9. A medical survey studied 40 students and found that 14 have blood group O and 6 have blood group AB. Find the probability that: (i) a randomly selected student has blood group O; (ii) a randomly selected student has blood group AB.
Answer: Given: Total number of students = 40, Number of students having blood group O = 14, Number of students having blood group AB = 6
(i) Required probability = \( \frac{14}{40} = 0.35 \)
(ii) Required probability = \( \frac{6}{40} = 0.15 \)

Exam Tip: Blood group probabilities vary by population - always use the sample data provided in the question rather than relying on general population statistics.

Question 10. In a class of 30 students, 6 students fall in the age interval 21 - 30 years. Find the probability that a randomly selected student lies in this age interval.
Answer: Given: Total number of students = 30, Number of students in the interval 21 - 30 = 6
Required probability = \( \frac{6}{30} = 0.2 \)

Exam Tip: Simplify the fraction 6/30 to 1/5 for easier understanding - this means 1 out of every 5 students falls into this age range.

Question 11. In a hospital, the ages of 360 patients are recorded. Find the probability that: (i) a randomly selected patient is between 30 and 40 years old; (ii) a randomly selected patient is between 50 and 70 years old; (iii) a randomly selected patient is less than 10 years old; (iv) a randomly selected patient is 10 years or older.
Answer: Given: Total number of patients = 360
(i) The probability of selecting a patient aged 30 years or more but less than 40 years: \( P = \frac{60}{360} = \frac{1}{6} \)
(ii) The probability of selecting a patient aged 50 years or more but less than 70 years: \( P = \frac{50 + 30}{360} = \frac{80}{360} = \frac{2}{9} \)
(iii) The probability of selecting a patient less than 10 years old: \( P = \frac{0}{360} = 0 \)
(iv) The probability of selecting a patient aged 10 years or older: \( P = \frac{360}{360} = 1 \)

Exam Tip: When an event is certain (everyone is 10+), its probability is 1; when an event is impossible (no one is under 10), its probability is 0 - these represent the extremes of the probability range.