RowQ
The Vault
RowQ
The Vault
CBSE Class 9 Maths · 11 questions · 26 marks
Probability puts a number on uncertainty, and at this stage that number comes from actually running the experiment and counting what happened. This is called empirical or experimental probability, and it is the reason surveys, quality checks and weather records can be turned into predictions. The whole chapter rests on one honest fraction: how often it happened, divided by how many times you tried.
A coin is tossed 500 times and a head appears 267 times. The empirical probability of getting a tail is:
Answer
The empirical probability of a tail is 0.466. Number of tails = total tosses - number of heads = 500 - 267 = 233. P(tail) = 233/500 = 0.466. (Check: P(head) = 267/500 = 0.534 and 0.466 + 0.534 = 1.)
Which of the following can never be the probability of an event?
Answer
The value -0.4 can never be a probability. Probability is a count of favourable trials divided by a count of total trials, so it can never be negative. The rule is 0 ≤ P(E) ≤ 1, and 0, 0.7 and 1 all lie in this range while -0.4 does not.
A die is thrown 300 times and the face 6 turns up 42 times. The empirical probability of getting a 6 is:
Answer
The empirical probability is 0.14. P(6) = (number of times 6 appeared)/(total throws) = 42/300. Dividing numerator and denominator by 6 gives 7/50 = 0.14. (The value 1/6 is what one might expect in theory, but empirical probability must be read off the actual data.)
If the probability of an event E is 3/8, then the probability of 'not E' is:
Answer
The probability of 'not E' is 5/8. Since E and 'not E' together cover every possible outcome, P(E) + P(not E) = 1. P(not E) = 1 - 3/8 = 8/8 - 3/8 = 5/8.
Assertion (A): If a coin is tossed 250 times and a head appears in every single toss, then the empirical probability of getting a tail in this experiment is 0. Reason (R): The empirical probability of any event is always either 0 or 1.
Answer
A is true but R is false. For Assertion A: number of tails = 250 - 250 = 0, so P(tail) = 0/250 = 0. Hence A is true. For Reason R: empirical probability can take any value from 0 to 1, such as 0.466 or 3/8. It is only 0 when the event never occurs and only 1 when it occurs in every trial, which are special cases and not a general rule. Hence R is false.
A quality inspector tests 1200 bulbs from a factory batch and finds 48 of them defective. Find the probability that a bulb picked at random from this batch is (a) defective, (b) not defective.
Answer
(a) P(defective) = 48/1200. Dividing numerator and denominator by 48 gives 1/25 = 0.04. (b) P(not defective) = 1 - 0.04 = 0.96. (Direct check: non-defective bulbs = 1200 - 48 = 1152, and 1152/1200 = 0.96.)
Two coins are tossed together 400 times. Two heads appeared 96 times, exactly one head appeared 210 times and no head appeared 94 times. Find the probability of each of these three events and verify that they add up to 1.
Answer
P(two heads) = 96/400 = 0.24. P(exactly one head) = 210/400 = 0.525. P(no head) = 94/400 = 0.235. Verification: 0.24 + 0.525 + 0.235 = 1. This is expected, because the three events are the elementary events of the experiment and every trial falls into exactly one of them: 96 + 210 + 94 = 400.
In a class of 45 students, the probability that a randomly chosen student wears spectacles is 4/15. How many students in the class wear spectacles, and how many do not?
Answer
Number wearing spectacles = P(E) × total number of students = (4/15) × 45. 45/15 = 3, so the number = 4 × 3 = 12 students. Number not wearing spectacles = 45 - 12 = 33 students. (Check: P(not wearing) = 33/45 = 11/15 = 1 - 4/15.)
A toll gate operator recorded the 800 vehicles that crossed one morning: Cars 340, two-wheelers 250, buses 90, trucks 120. For a vehicle chosen at random from this record, find the probability that it is (a) a car, (b) a two-wheeler, (c) not a truck, (d) a bus or a truck. (e) Verify that the probabilities of the four vehicle types add up to 1.
Answer
Total number of trials (vehicles) = 340 + 250 + 90 + 120 = 800. (a) P(car) = 340/800 = 17/40 = 0.425. (b) P(two-wheeler) = 250/800 = 5/16 = 0.3125. (c) Vehicles that are not trucks = 800 - 120 = 680. P(not a truck) = 680/800 = 17/20 = 0.85. (Alternatively, P(truck) = 120/800 = 0.15 and 1 - 0.15 = 0.85.) (d) Buses and trucks together = 90 + 120 = 210. P(bus or truck) = 210/800 = 21/80 = 0.2625. (e) P(car) + P(two-wheeler) + P(bus) + P(truck) = 0.425 + 0.3125 + 0.1125 + 0.15 = 1, where P(bus) = 90/800 = 0.1125 and P(truck) = 0.15. The sum is 1 because every vehicle recorded belongs to exactly one of the four types.
A die was thrown 500 times and the frequency of each face was recorded: Face: 1, 2, 3, 4, 5, 6 Frequency: 78, 85, 80, 91, 84, 82 (a) Find the probability of getting a number greater than 4. (b) Find the probability of getting an even number. (c) Show that the probabilities of the six faces add up to 1. (d) If the die is thrown 2000 more times under the same conditions, about how many times would you expect the face 4 to appear?
Answer
First check the total: 78 + 85 + 80 + 91 + 84 + 82 = 500, matching the number of throws. (a) Numbers greater than 4 are 5 and 6, with frequencies 84 and 82. Favourable trials = 84 + 82 = 166. P(number greater than 4) = 166/500 = 83/250 = 0.332. (b) Even numbers are 2, 4 and 6, with frequencies 85, 91 and 82. Favourable trials = 85 + 91 + 82 = 258. P(even number) = 258/500 = 129/250 = 0.516. (c) The six probabilities are 78/500, 85/500, 80/500, 91/500, 84/500 and 82/500. Adding them gives (78 + 85 + 80 + 91 + 84 + 82)/500 = 500/500 = 1. This must happen because each throw produces exactly one of the six faces. (d) P(face 4) = 91/500 = 0.182. Expected frequency in 2000 throws ≈ 0.182 × 2000 = 364 times. This is an estimate, not a guarantee, since empirical probability only predicts the long-run pattern.
Read the following and answer the questions that follow: A juice stall near a school recorded the choice of flavour made by 250 customers during one week: Mango 90, Orange 65, Guava 45, Lemon 50. The owner wants to use this record to plan next week's stock. (a) Find the probability that a customer chosen at random from this record ordered mango. (b) Find the probability that a customer did not order orange. (c) Which flavour was the least likely to be ordered, and what is its probability? (d) If 600 customers visit next week, roughly how many mango juices should the owner prepare for?
Answer
Total customers = 90 + 65 + 45 + 50 = 250. (a) P(mango) = 90/250 = 9/25 = 0.36. (b) P(orange) = 65/250 = 13/50 = 0.26. P(did not order orange) = 1 - 0.26 = 0.74. (Direct check: 250 - 65 = 185 and 185/250 = 0.74.) (c) Guava has the smallest frequency, 45, so it was least likely. P(guava) = 45/250 = 9/50 = 0.18. (d) Expected mango orders = P(mango) × 600 = 0.36 × 600 = 216 juices. The owner should plan for about 216 mango juices, keeping in mind that this is an estimate based on last week's pattern.
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