Probability: Questions and Answers

42 probability questions solved step by step. Open a question for the full working, the marks for each step and exam practice.

• Can we predict these outcomes with 100% certainty?Answer: No. We know all the possible outcomes (rain or no rain; win, draw or lose; which student is chosen), but we cannot know in advance which one will happen. We can only say how likely each one is.
• Such unpredictability can be useful sometimes! For example, in a cricket match, the fact that a coin is tossed to decide which team will bat first is considered to be a fair method. Can you explain why?Answer: A fair coin has two equally likely outcomes, heads and tails, each with probability 12. Nobody can predict or control the result, so each captain has exactly the same chance (12) of winning the toss. Neither team is favoured, so the method is fair.
• Have you wondered what makes an event like rain random?Answer: Rain depends on many complex conditions in the atmosphere (temperature, humidity, wind patterns, pressure) and is very sensitive to them, so it is impossible to predict with total certainty. We can only estimate the probability of rain from patterns in past data.
• Ask your friend to predict the outcome of a ₹ 1 coin you toss. Do you see that your friend could guess heads or tails but could not know for certain? That’s randomness! All possible results are known, but each individual try is unpredictable.Answer: Yes. Your friend knows the coin will show heads or tails, but cannot know which. Over many tosses the guesses come out right only about half the time (probability 12 each time), which shows each toss is unpredictable: that is randomness.
• Rank the following events on a scale from 0 (Impossible) to 1 (Certain). Label each event: Impossible, less likely, equally likely (even chance), more likely, certain. Give reasons why you gave each event its ranking.Answer: From 0 to 1: (ii) snow in Mumbai in July: impossible (0); (iii) an elephant in your classroom: less likely (very close to 0); (iv) greeting at least one friend tomorrow: more likely (close to 1); (i) Monday after Sunday: certain (1).
• The set of all possible outcomes in an experiment is called the sample space. The sample space may be listed within brackets, separated by commas as shown below.Answer: (i) Coin tossed: sample space {H, T}. (ii) Die rolled: sample space {1, 2, 3, 4, 5, 6}.
• Suppose you roll a die 50 times, and it lands on a 4 exactly 8 times.Answer: Experimental probability of rolling a 4 = 850 = 0.16 or 16%. This is also the relative frequency of rolling a 4.
• If you roll a standard 6-sided die, what is the theoretical probability of getting a 4?Answer: P(rolling a 4) = 16 = 0.1666… ≈ 0.167 or 16.7%.
• A letter is picked at random from the word ‘PROBABILITY’. What is the probability of picking the letter B?Answer: P(picking the letter B) = 211 = 0.1818… ≈ 0.182 or 18.2%.
• Suppose you anonymously collect information regarding the favourite fruit of 50 students in your class. Let us assume that the results are: 20 students like mango, 15 students like apples, 10 students like bananas, and 5 students like grapes. Let us play a game! Suppose we randomly pick one student from the class and try to guess their favourite fruit. What’s the probability that the student’s favourite fruit is mango?Answer: P(mango) = 2050 = 0.4, a 40% chance. Used for a school of 1500 students, this estimate means buying about 600 mangoes (40% of 1500).
• If I have rolled a 4 on a die 8 times in succession, the probability of rolling a 4 again is still only ≈ 0.16 (assuming the die is fair). Probability does not tell you what will happen next but predicts what will happen in the long run.Answer: True. A fair die has no memory, so on every roll P(4) = 16 = 0.1666… (about 0.16 to 0.17), however many 4s came before. Probability describes the long run: in 600 rolls you expect about 100 fours, but it cannot tell you the next roll.
• Let us say you are playing Snakes and Ladders, and you are rolling a fair 6-sided die to move. You have just rolled the die three times in a row, and each time you got a 6. Now, you think: ‘I have already rolled three 6s — there is no way I will get a 6 again on the next roll!’Answer: The thinking is wrong; it is the Gambler’s Fallacy. Each roll is independent, so the probability of a 6 on the next roll is still 16 ≈ 0.167 (about 16.7%).
• A teacher mixes a large bag of sweets of different colours and randomly selects a sample of 30 sweets. She counts the number of sweets of each colour:
  10 red sweets | 8 green sweets | 7 yellow sweets | 5 blue sweetsAnswer: (i) P(green) = 830 = 415 ≈ 0.27. (ii) About 730 × 600 = 140 yellow sweets.
• A survey is conducted at a school where a random sample of 40 students is asked about their favourite club. The responses are:
  14 students: Science Club | 11 students: Arts Club |
  9 students: Sports Club | 6 students: Debate Club
  Assume there are 800 students in the whole school.Answer: (i) P(Arts Club) = 1140 = 0.275. (ii) About 940 × 800 = 180 students prefer the Sports Club.
• Toss a coin 20 times and record the result each time (heads or tails).Answer: Your counts depend on your own tosses. In one sample run: (i) 11 heads (ii) 9 tails (iii) experimental P(heads) = 1120 = 0.55 (iv) P(tails) on the next toss = 12, whatever happened before.
• Toss a paper cup into the air 100 times. After each toss record whether the cup lands on its bottom, upside down on its top or on its side (See the figure). Assign probabilities to the outcomes by using experimental probability.Answer: Use P = count100 for each landing position. One sample set of results: bottom 12, top 23, side 65, giving P(bottom) = 0.12, P(top) = 0.23, P(side) = 0.65 (total 1). Your own results will differ, but the cup usually lands on its side most often.
• What is the probability of getting an even number when rolling a fair 6-sided die?Answer: P(even) = 36 = 12 = 0.5.
• Suppose you roll a 6-sided die 12 times and get a ‘3’ three times.Answer: (i) Experimental P(3) = 312 = 14 = 0.25. (ii) Theoretical P(3) = 16 ≈ 0.167. (iii) 12 rolls is a small number of trials, so chance has a big effect. With 60, 600, 6000 rolls the experimental probability gets closer and closer to 16 (about 10, 100, 1000 threes).
• When we used the sample space {Rain, No Rain}, we focused only on whether it will rain or not. However, if we want to include different amounts of rainfall like drizzle, light rain or heavy rain, we need to expand the sample space to {No Rain, Drizzle, Light Rain, Heavy Rain} so that it better matches the level of detail required for the question. It is important to ensure the sample space is detailed enough to suit the specific problem being studied.Answer: The same situation can have different sample spaces. {Rain, No Rain} (2 outcomes) only answers “will it rain?”; {No Rain, Drizzle, Light Rain, Heavy Rain} (4 outcomes) answers “how much will it rain?”. Choose the sample space whose outcomes match the question, and make sure it still lists every possibility once.
• When a single 6-sided die is rolled, what is the total number of possible outcomes in the sample space?Answer: S = {1, 2, 3, 4, 5, 6}, so n(S) = 6.
• For the following experiments write down the sample space S.Answer: (i) S = {1H, 1T, 2H, 2T, 3H, 3T, 4H, 4T, 5H, 5T, 6H, 6T}, 12 outcomes. (ii) S = {−5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5}, 11 outcomes. (iii) S = {G₁, G₂, G₃, G₄, G₅, R₁, R₂, …, R₇}, 12 outcomes (by colour only: {Green, Red}).
• In a village fair, there are 3 popular snacks available: Samosa, Pakora, and Bhaji. For drinks, villagers can choose either Chai or Lassi.Answer: (i) S = {(Samosa, Chai), (Samosa, Lassi), (Pakora, Chai), (Pakora, Lassi), (Bhaji, Chai), (Bhaji, Lassi)}: 6 combinations. (ii) E = {(Samosa, Chai), (Samosa, Lassi)}.
• Experiment: Toss a fair coin two times.Answer: The tree diagram shows 4 possible outcomes: S = {HH, HT, TH, TT}. Each has probability 14, so P(HH) = 14 = 0.25 or 25%.
• Can you calculate the probability of getting one head and one tail?Answer: When a fair coin is tossed twice, ‘one head and one tail’ = {HT, TH}: 2 of the 4 equally likely outcomes. P = 24 = 12 = 0.5.
• There are two fruit baskets A and B. Basket A has one apple and two oranges. Basket B has one banana and one mango. You randomly pick one fruit from each basket.Answer: (i) Tree: 3 branches for basket A (Apple, Orange 1, Orange 2), each splitting into Banana and Mango: 6 paths. (ii) S = {(Apple, Banana), (Apple, Mango), (Orange 1, Banana), (Orange 1, Mango), (Orange 2, Banana), (Orange 2, Mango)}. (iii) P(apple and banana) = 16.
• Let us say that you have a box containing 3 red pens, 4 black pens and 2 green pens. You pick a pen (without looking) from the box and put it back. Then your friend does the same.Answer: (i) 9 colour outcomes: RR, RB, RG, BR, BB, BG, GR, GB, GG (R = red, B = black, G = green; your pen first). (ii) P(same colour) = 981 + 1681 + 481 = 2981 ≈ 0.36.
• Fill in the blanks.Answer: (i) 0 (ii) sample space (iii) 1 (iv) 12 (0.5)
• In a survey of 50 students, 15 students said they liked football. The number of students who like football is 15, and the ________ (frequency/relative frequency) is __________ (fill in the fraction or decimal).Answer: The relative frequency is 1550 = 310 = 0.3.
• Which of the following experiments have equally likely outcomes? Explain.Answer: Equally likely: (ii), (iii) and (v). Not equally likely: (i) (a car usually starts) and (iv) (blue is more likely: 710 against 310).
• Write the sample space and calculate the probability based on the given information.Answer: (i) 34 (ii) 12 (iii) 13 (iv) 12 (v) 38
• A bag has 3 candies: strawberry, lemon, and mint. One is picked at random. What is the probability of picking a strawberry candy?Answer: P(strawberry) = 13 ≈ 0.33.
• A child has 2 shirts (one red and one blue) and 3 types of pants (jeans, khakis, and shorts). List all the possible combinations of outfits consisting of one shirt and one pair of pants. Display your answer in a table format.Answer: 2 × 3 = 6 outfits: (Red, Jeans), (Red, Khakis), (Red, Shorts), (Blue, Jeans), (Blue, Khakis), (Blue, Shorts).
• A tyre company records distances before replacement in 1000 cases. Find the probability that a randomly chosen tyre lasts:Answer: (i) 201000 = 0.02 (ii) 5351000 = 0.535 (iii) 4451000 = 0.445
• The letters of the word ‘PEACE’ are placed on cards. Leela draws a card without looking.Answer: (i) P(P, E or C) = 45 (ii) P(not E) = 35
• A game of chance consists of spinning an arrow (see the figure.) which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, and these are equally likely outcomes. What is the probability that it will point atAnswer: (i) 18 (ii) 12 (iii) 34 (iv) 1 (v) 14
• A basket contains 4 red balls and 5 blue balls. One ball is drawn and laid aside, and a second ball is drawn. Draw a tree diagram to represent the possible outcomes and probabilities. Use the tree diagram to answer the following questions.Answer: Tree: first draw R 49, B 59; second draw after R: R 38, B 58; after B: R 48, B 48. (i) P(red then blue) = 49 × 58 = 518. (ii) P(2 blue) = 59 × 48 = 518.
• I throw a pair of 6-sided dice. Write down an event that has a probability of 0 and an outcome that has a probability of 1.Answer: Probability 0: “the sum of the two numbers is 13” (impossible; the largest sum is 12). Probability 1: “the sum of the two numbers is between 2 and 12” (certain; it always is).
• Write the sample space and calculate the probability based on the given information.Answer: (i) 836 = 29 (ii) 5272 = 1318 (iii) 28 = 14 (iv) 1224 = 12 (v) 964
• A box contains 4 balls numbered 1 to 4. Record a sample space using a tree diagram for the following experiments:Answer: (i) With replacement: 4 × 4 = 16 outcomes, (1, 1) to (4, 4). (ii) Without replacement: 4 × 3 = 12 outcomes (no repeated number). (iii) n(S) = 16 and 12.
• List the elements of a sample space for the simultaneous tossing of a coin and drawing of a card from a set of 6 cards numbered 1 through 6.Answer: S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}; n(S) = 2 × 6 = 12.
• Three coins are tossed, and the number of heads is recorded. Which of the following lists is a sample space for this experiment? Why do the other lists fail to qualify as a sample space?Answer: (iv) {0, 1, 2, 3} is the sample space. (i) leaves out 0 (TTT), (ii) leaves out 3 (HHH), and (iii) includes 4, which is impossible with 3 coins.
• Suppose you drop a dye at random on the rectangular region shown in the figure. What is the probability that it will land inside the circle with a diameter of 1 m?Answer: P = Area of circleArea of rectangle = π × 0.5²3 × 2 = π24 ≈ 0.13 (about 13%).

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