The Mathematics of Maybe: Introduction to Probability

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 coin toss is considered fair because it has only two possible outcomes — Head and Tail — and both are equally likely, each with a probability of $\dfrac{1}{2}$.

Since neither team can predict or control which outcome will turn up, and both outcomes have the same chance, no team is given any advantage over the other. The result depends purely on chance.

This is why tossing a coin is accepted as an unbiased and fair way to decide which team will bat first.

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. Before the coin is tossed, we already know the complete list of possible outcomes — Head and Tail. So a friend can confidently guess one of them.

However, the friend cannot know for certain which one will actually come up on a single toss, because the result is decided by chance.

This is exactly what randomness means: the set of all possible outcomes is known in advance, but the outcome of any individual trial cannot be predicted with certainty.

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.

(i) The next Monday will come after Sunday.

(ii) It will snow in Mumbai in July.

(iii) An elephant will walk through your classroom today.

(iv) You will greet at least one friend at school tomorrow.

Answer

(i) The next Monday will come after Sunday.

In the weekly calendar, Monday always follows Sunday. Therefore, this event is certain.

∴ The event is Certain (probability = 1).

(ii) It will snow in Mumbai in July.

Mumbai is a hot, coastal city, and July falls in its monsoon (rainy) season with high temperatures. Snowfall in such conditions never happens.

∴ The event is Impossible (probability = 0).

(iii) An elephant will walk through your classroom today.

An elephant has no reason to enter a classroom, and in normal circumstances this simply does not happen.

∴ The event is Impossible (probability = 0).

(iv) You will greet at least one friend at school tomorrow.

If you go to school, you will almost surely meet and greet at least one of your friends, though it is not absolutely guaranteed.

∴ The event is More likely (probability close to 1).

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

Yes, this is correct. Each roll of a fair die is an independent event — the die has no memory of previous rolls.

So even after getting a 4 eight times in a row, the probability of getting a 4 on the next roll is still:

$\dfrac{1}{6} \approx 0.16$

The chance does not increase or decrease because of past results.

Probability cannot tell us the outcome of the next single roll. It only tells us what to expect in the long run — for example, over a very large number of rolls, a 4 will appear in about $\dfrac{1}{6}$ of them.

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 sweets

(i) Calculate the probability that a randomly picked sweet from the sample is green.

(ii) If there are 600 sweets in total in the large bag, estimate how many are likely to be yellow, based on the sample results.

Answer

Total number of sweets in the sample = 30.

(i) Number of green sweets = 8.

P(picking a green sweet) = $\dfrac{\text{number of green sweets}}{\text{total number of sweets in the sample}}$

$= \dfrac{8}{30} \\[1em] = \dfrac{4}{15}$

∴ Probability of picking a green sweet = $\mathbf{\dfrac{4}{15}}$

(ii) Number of yellow sweets in the sample = 7.

P (picking a yellow sweet) = $\dfrac{7}{30}$

Estimated number of yellow sweets in the bag of 600

$= \dfrac{7}{30} \times 600 \\[1em] = 7 \times 20 \\[1em] = 140$

∴ About 140 sweets in the bag are likely to be yellow.

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.

(i) What is the probability that a randomly chosen student from the sample prefers the Arts Club?

(ii) Using the sample results, estimate how many students in the whole school are likely to prefer the Sports Club.

Answer

Total number of students in the sample = 40.

(i) Number of students who prefer the Arts Club = 11.

P (student prefers the Arts Club) = $\dfrac{\text{number who prefer the Arts Club}}{\text{total number of students in the sample}}$

= $\dfrac{11}{40}$

∴ Probability that a student prefers the Arts Club = $\mathbf{\dfrac{11}{40}}$

(ii) Number of students who prefer the Sports Club in the sample = 9.

P (student prefers the Sports Club) = $\dfrac{9}{40}$

Estimated number of students in the whole school of 800

$= \dfrac{9}{40} \times 800 \\[1em] = 9 \times 20 \\[1em] = 180$

∴ About 180 students in the whole school are likely to prefer the Sports Club.

Toss a coin 20 times and record the result each time (heads or tails).

(i) How many times did you get heads?

(ii) How many times did you get tails?

(iii) Calculate the experimental probability of getting heads.

(iv) If you toss the coin once more, what is the probability of getting tails?

Answer

This is an activity, so the exact numbers depend on your own tosses. A sample set of results is used below to show the method.

Total number of trials = 20.

(i) Suppose Heads was obtained = 11 times.

∴ Number of heads = 11

(ii) Then Tails was obtained = 20 − 11 = 9 times.

∴ Number of tails = 9

(iii) Experimental P (getting heads) = $\dfrac{\text{number of heads obtained}}{\text{total number of trials}}$

= $\dfrac{11}{20}$

∴ Experimental probability of getting heads = $\mathbf{\dfrac{11}{20}}$ (= 0.55)

(iv) The coin is fair, and each toss is independent of the previous tosses. So for one more toss, we use the theoretical probability.

P (getting tails) = $\dfrac{\text{number of favourable outcomes}}{\text{number of possible outcomes}}$

= $\dfrac{1}{2}$

∴ Probability of getting tails on one more toss = $\mathbf{\dfrac{1}{2}}$

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 Fig. 7.5). Assign probabilities to the outcomes by using experimental probability.

Answer

This is an activity, so the exact numbers depend on your own tosses. A sample set of results is used below to show how the experimental probabilities are assigned.

Total number of trials = 100.

Suppose the cup landed as follows :

Bottom : 20 times; Top : 16 times; Side : 64 times.

P(lands on its bottom) = $\dfrac{\text{number of times it lands on bottom}}{\text{total number of trials}}$

$= \dfrac{20}{100} \\[1em] = \dfrac{1}{5}$

P(lands on its top) = $\dfrac{16}{100} = \dfrac{4}{25}$

P(lands on its side) = $\dfrac{64}{100} = \dfrac{16}{25}$

∴ P(bottom) = $\mathbf{\dfrac{1}{5}}$, P(top) = $\mathbf{\dfrac{4}{25}}$, P(side) = $\mathbf{\dfrac{16}{25}}$

What is the probability of getting an even number when rolling a fair 6-sided die?

Answer

When a fair 6-sided die is rolled, the sample space is S = {1, 2, 3, 4, 5, 6}.

Number of possible outcomes = 6.

The even numbers are 2, 4 and 6.

Number of favourable outcomes = 3.

P(getting an even number) = $\dfrac{\text{number of favourable outcomes}}{\text{number of possible outcomes}}$

$= \dfrac{3}{6} \\[1em] = \dfrac{1}{2}$

∴ Probability of getting an even number = $\mathbf{\dfrac{1}{2}}$

Suppose you roll a 6-sided die 12 times and get a '3' three times.

(i) What is the experimental probability of rolling a '3'?

(ii) What is the theoretical probability of rolling a '3'?

(iii) Why might these probabilities be different? What would you expect to happen if you roll the die 60, 600, or 6000 times?

Answer

(i) Total number of trials = 12, and a '3' was obtained 3 times.

Experimental P(rolling a 3) = $\dfrac{\text{number of times 3 occurred}}{\text{total number of trials}}$

$= \dfrac{3}{12} \\[1em] = \dfrac{1}{4}$

∴ Experimental probability of rolling a 3 = $\mathbf{\dfrac{1}{4}}$

(ii) A standard die has 6 equally likely faces, and only one of them is a '3'.

Theoretical P(rolling a 3) = $\dfrac{\text{number of favourable outcomes}}{\text{number of possible outcomes}}$

= $\dfrac{1}{6}$

∴ Theoretical probability of rolling a 3 = $\dfrac{1}{6}$

(iii) The two probabilities are different because the experimental probability is based on only a small number of trials (12). With few trials, the observed result can easily differ from the ideal value.

As the number of trials increases to 60, 600 and then 6000, the experimental probability tends to get closer and closer to the theoretical probability of $\dfrac{1}{6}$. This is known as the Law of Large Numbers.

When we used the sample space {Rain, No Rain} in Example 1, 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

Yes. The sample space should always be chosen to match the level of detail that the question requires.

If we only want to know whether it rains, the sample space S = {Rain, No Rain} is enough.

But if we want to study how much it rains, this sample space is too simple. We must expand it to S = {No Rain, Drizzle, Light Rain, Heavy Rain} so that every outcome of interest is included.

So, the same experiment can have different sample spaces depending on the question being asked. The sample space must be detailed enough to describe all the outcomes that matter for the specific problem.

When a single 6-sided die is rolled, what is the total number of possible outcomes in the sample space?

Answer

When a single 6-sided die is rolled, the top face can show any one of the numbers 1, 2, 3, 4, 5 or 6.

So the sample space is S = {1, 2, 3, 4, 5, 6}.

∴ The total number of possible outcomes, n(S) = 6.

For the following experiments write down the sample space S.

(i) Rolling a die and tossing a coin together.

(ii) Choosing a random integer between –5 and +5.

(iii) A box containing 5 green and 7 red balls. One ball is drawn at random.

Answer

(i) Rolling a die and tossing a coin together.

The die can show 1, 2, 3, 4, 5 or 6, and the coin can show Head (H) or Tail (T). Pairing each die result with each coin result :

S = {(1, H), (2, H), (3, H), (4, H), (5, H), (6, H), (1, T), (2, T), (3, T), (4, T), (5, T), (6, T)}

∴ n(S) = 12

(ii) Choosing a random integer between –5 and +5.

The integers lying between –5 and +5 are :

S = {–4, –3, –2, –1, 0, 1, 2, 3, 4}

∴ n(S) = 9

(iii) A box containing 5 green and 7 red balls, one ball drawn at random.

The drawn ball can only be either green or red.

S = {Green, Red}

∴ n(S) = 2

In a village fair, there are 3 popular snacks available: Samosa, Pakora, and Bhaji. For drinks, villagers can choose either Chai or Lassi.

(i) List the sample space of all possible snack and drink combinations a person could choose at the fair.

(ii) List the event 'Selecting Samosa as a snack.'

Answer

Snacks: Samosa, Pakora, Bhaji.

Drinks: Chai, Lassi.

(i) Pairing each snack with each drink, the sample space is:

S = {(Samosa, Chai), (Samosa, Lassi), (Pakora, Chai), (Pakora, Lassi), (Bhaji, Chai), (Bhaji, Lassi)}

∴ n(S) = 6

(ii) The event 'Selecting Samosa as a snack' contains all combinations in which the snack is Samosa:

E = {(Samosa, Chai), (Samosa, Lassi)}

Toss a fair coin two times. Can you calculate the probability of getting one head and one tail?

Answer

When a fair coin is tossed two times, the sample space is:

S = {HH, HT, TH, TT}

Number of possible outcomes, n(S) = 4.

The outcomes with one head and one tail are HT and TH.

Number of favourable outcomes = 2.

P(one head and one tail) = $\dfrac{\text{number of favourable outcomes}}{\text{number of possible outcomes}}$

$= \dfrac{2}{4} \\[1em] = \dfrac{1}{2}$

∴ Probability of getting one head and one tail = $\mathbf{\dfrac{1}{2}}$

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.

(i) Draw a tree diagram showing all possible pairs of fruits.

(ii) List the sample space.

(iii) What is the probability of picking one apple and one banana?

Answer

Basket A has 3 fruits: Apple (A), Orange (O₁) and Orange (O₂).

Basket B has 2 fruits: Banana (B) and Mango (M).

(i) Drawing a line from a starting point to each fruit of Basket A, and from each of those to each fruit of Basket B, gives the tree diagram:

(ii) Pairing each fruit of Basket A with each fruit of Basket B, the sample space is:

S = {(A, B), (A, M), (O₁, B), (O₁, M), (O₂, B), (O₂, M)}

∴ n(S) = 6

(iii) The favourable outcome 'one apple and one banana' is (A, B), which occurs only once.

P(one apple and one banana) = $\dfrac{\text{number of favourable outcomes}}{\text{number of possible outcomes}}$

= $\dfrac{1}{6}$

∴ Probability of picking one apple and one banana = $\mathbf{\dfrac{1}{6}}$

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.

(i) What are the possible outcomes of the pen colours? Can you draw a tree diagram representing the possible outcomes?

(ii) Can you use the tree diagram to guess the probability that both you and your friend pick pens of the same colour?

Answer

Total number of pens = 3 + 4 + 2 = 9.

Since the first pen is put back before the second is picked, both picks are independent.

P(Red) = $\dfrac{3}{9}$ = $\dfrac{1}{3}$, P(Black) = $\dfrac{4}{9}$, P(Green) = $\dfrac{2}{9}$.

(i) Each person can pick a Red (R), Black (Bl) or Green (G) pen. Pairing your colour with your friend's colour gives 3 × 3 = 9 possible outcomes:

S = {R-R, R-Bl, R-G, Bl-R, Bl-Bl, Bl-G, G-R, G-Bl, G-G}

The tree diagram representing these outcomes is :

(ii) 'Both pick pens of the same colour' means both Red, or both Black, or both Green.

P(both same colour) = P(both red) + P(both black) + P(both green)

$= \left(\dfrac{3}{9}\right)^2 + \left(\dfrac{4}{9}\right)^2 + \left(\dfrac{2}{9}\right)^2 \\[1em] = \dfrac{9}{81} + \dfrac{16}{81} + \dfrac{4}{81} \\[1em] = \dfrac{29}{81}$

∴ Probability that both pick pens of the same colour = $\mathbf{\dfrac{29}{81}}$

Fill in the blanks.

(i) The probability of an impossible event is ............... .

(ii) The set of all possible outcomes of a random experiment is called the ............... .

(iii) The probability of an event that is certain to happen is ............... .

(iv) Tossing a fair coin has a probability of ............... for getting heads.

Answer

(i) The probability of an impossible event is 0.

(ii) The set of all possible outcomes of a random experiment is called the sample space.

(iii) The probability of an event that is certain to happen is 1.

(iv) Tossing a fair coin has a probability of $\mathbf{\dfrac{1}{2}}$ for getting heads.

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 number of students who like football, 15, is the frequency.

The relative frequency is found by dividing this frequency by the total number of students.

Relative frequency = $\dfrac{\text{number of students who like football}}{\text{total number of students}}$

$= \dfrac{15}{50} \\[1em] = \dfrac{3}{10} \\[1em] = 0.3$

∴ The number 15 is the frequency, and the relative frequency is $\mathbf{\dfrac{3}{10}}$ (= 0.3).

Which of the following experiments have equally likely outcomes? Explain.

(i) A driver attempts to start a car. The car starts or does not start.

(ii) Tossing a fair coin once.

(iii) Rolling a fair 6-sided die.

(iv) Choosing a marble randomly from a bag that contains 3 red marbles and 7 blue marbles.

(v) A baby is born. It is a boy or a girl.

Answer

(i) A driver attempts to start a car.

The two outcomes 'starts' and 'does not start' are not equally likely, because a car in good condition is far more likely to start than not.

(ii) Tossing a fair coin once.

Head and Tail are equally likely, since a fair coin is symmetrical and each outcome has probability $\dfrac{1}{2}$.

(iii) Rolling a fair 6-sided die.

The six faces are equally likely, since a fair die is symmetrical and each face has probability $\dfrac{1}{6}$.

(iv) Choosing a marble from a bag of 3 red and 7 blue marbles.

The outcomes 'red' and 'blue' are not equally likely, because there are more blue marbles, so choosing blue is more likely than red.

(v) A baby is born — boy or girl.

The two outcomes are taken to be (approximately) equally likely, since a baby is about equally likely to be a boy or a girl.

Write the sample space and calculate the probability based on the given information.

(i) Two coins are tossed at the same time. What is the probability of getting at least one head?

(ii) Ten identical cards numbered 1 to 10 are placed in a box. One card is drawn at random. What is the probability of drawing a card with an even number?

(iii) A die is rolled once. What is the probability of getting a number greater than 4?

(iv) A bag contains 3 red balls, 2 blue balls, and 1 green ball. One ball is picked at random. What is the probability that it is not red?

(v) Three coins are tossed simultaneously. What is the probability of getting exactly two heads?

Answer

(i) Two coins tossed at the same time.

Sample space S = {HH, HT, TH, TT}, so n(S) = 4.

'At least one head' = {HH, HT, TH}, which has 3 outcomes.

P(at least one head) = $\dfrac{3}{4}$

∴ Probability of getting at least one head = $\mathbf{\dfrac{3}{4}}$

(ii) Ten cards numbered 1 to 10.

Sample space S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, so n(S) = 10.

Even numbers = {2, 4, 6, 8, 10}, which has 5 outcomes.

P(even number) = $\dfrac{5}{10}$ = $\dfrac{1}{2}$

∴ Probability of drawing an even-numbered card = $\mathbf{\dfrac{1}{2}}$

(iii) A die rolled once.

Sample space S = {1, 2, 3, 4, 5, 6}, so n(S) = 6.

Numbers greater than 4 = {5, 6}, which has 2 outcomes.

P(number greater than 4) = $\dfrac{2}{6}$ = $\dfrac{1}{3}$

∴ Probability of getting a number greater than 4 = $\mathbf{\dfrac{1}{3}}$

(iv) Bag with 3 red, 2 blue and 1 green ball.

Total balls = 3 + 2 + 1 = 6, so n(S) = 6.

'Not red' means blue or green = 2 + 1 = 3 balls.

P (not red) = $\dfrac{3}{6}$ = $\dfrac{1}{2}$

∴ Probability that the ball is not red = $\mathbf{\dfrac{1}{2}}$

(v) Three coins tossed simultaneously.

Sample space S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}, so n(S) = 8.

'Exactly two heads' = {HHT, HTH, THH}, which has 3 outcomes.

P (exactly two heads) = $\dfrac{3}{8}$

∴ Probability of getting exactly two heads = $\mathbf{\dfrac{3}{8}}$

A bag has 3 candies: strawberry, lemon, and mint. One is picked at random. What is the probability of picking a strawberry candy?

Answer

Sample space S = {Strawberry, Lemon, Mint}, so n(S) = 3.

The favourable outcome (strawberry) occurs once.

P(strawberry candy) = $\dfrac{\text{number of favourable outcomes}}{\text{number of possible outcomes}}$

= $\dfrac{1}{3}$

∴ Probability of picking a strawberry candy = $\mathbf{\dfrac{1}{3}}$

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

Each shirt can be paired with each type of pants, giving 2 × 3 = 6 possible outfits.

∴ The 6 possible outfits are: (Red, Jeans), (Red, Khakis), (Red, Shorts), (Blue, Jeans), (Blue, Khakis) and (Blue, Shorts).

A tyre company records distances before replacement in 1000 cases.

Find the probability that a randomly chosen tyre lasts:

(i) Less than 4000 km.

(ii) Between 4000 and 14000 km.

(iii) More than 14000 km.

Answer

Total number of cases = 20 + 210 + 325 + 445 = 1000.

(i) Number of tyres lasting less than 4000 km = 20.

P(less than 4000 km) = $\dfrac{\text{number of such cases}}{\text{total number of cases}}$

$= \dfrac{20}{1000} \\[1em] = \dfrac{1}{50}$

∴ Probability that a tyre lasts less than 4000 km = $\mathbf{\dfrac{1}{50}}$

(ii) Number of tyres lasting between 4000 and 14000 km = 210 + 325 = 535.

P(between 4000 and 14000 km) = $\dfrac{535}{1000}$

$= \dfrac{535}{1000} \\[1em] = \dfrac{107}{200}$

∴ Probability that a tyre lasts between 4000 and 14000 km = $\mathbf{\dfrac{107}{200}}$

(iii) Number of tyres lasting more than 14000 km = 445.

P(more than 14000 km) = $\dfrac{445}{1000}$

$= \dfrac{445}{1000} \\[1em] = \dfrac{89}{200}$

∴ Probability that a tyre lasts more than 14000 km = $\mathbf{\dfrac{89}{200}}$

The letters of the word 'PEACE' are placed on cards. Leela draws a card without looking.

(i) What is the probability that it is a P, E or C?

(ii) What is the probability that it is not an E?

Answer

The word 'PEACE' has 5 letters: P, E, A, C, E.

Total number of cards, n(S) = 5.

(i) The cards bearing P, E or C are: P (1 card), E (2 cards) and C (1 card).

Number of favourable cards = 1 + 2 + 1 = 4.

P(a P, E or C) = $\dfrac{4}{5}$

∴ Probability that it is a P, E or C = $\mathbf{\dfrac{4}{5}}$

(ii) The cards that are 'not an E' are : P, A and C.

Number of favourable cards = 5 − 2 = 3.

P(not an E) = $\dfrac{3}{5}$

∴ Probability that it is not an E = $\mathbf{\dfrac{3}{5}}$

A game of chance consists of spinning an arrow (see Fig. 7.7.) 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 at

(i) 8?

(ii) An odd number?

(iii) A number greater than 2?

(iv) A number less than 9?

(v) A multiple of 3?

Answer

The arrow can point at any one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, and all are equally likely.

Number of possible outcomes, n(S) = 8.

(i) Pointing at 8: favourable outcome = {8}, that is 1 outcome.

P(8) = $\dfrac{1}{8}$

∴ Probability of pointing at 8 = $\mathbf{\dfrac{1}{8}}$

(ii) Odd numbers = {1, 3, 5, 7}, that is 4 outcomes.

P(an odd number) = $\dfrac{4}{8} = \dfrac{1}{2}$

∴ Probability of pointing at an odd number = $\mathbf{\dfrac{1}{2}}$

(iii) Numbers greater than 2 = {3, 4, 5, 6, 7, 8}, that is 6 outcomes.

P(greater than 2) = $\dfrac{6}{8} = \dfrac{3}{4}$

∴ Probability of pointing at a number greater than 2 = $\mathbf{\dfrac{3}{4}}$

(iv) Numbers less than 9 = {1, 2, 3, 4, 5, 6, 7, 8}, that is all 8 outcomes.

P(less than 9) = $\dfrac{8}{8}$ = 1

∴ Probability of pointing at a number less than 9 = 1

(v) Multiples of 3 = {3, 6}, that is 2 outcomes.

P(a multiple of 3) = $\dfrac{2}{8} = \dfrac{1}{4}$

∴ Probability of pointing at a multiple of 3 = $\mathbf{\dfrac{1}{4}}$

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.

(i) What is the probability of drawing a red ball and then a blue ball?

(ii) What is the probability of drawing 2 blue balls?

Answer

Total number of balls = 4 red + 5 blue = 9.

Since the first ball is laid aside (not replaced), only 8 balls remain for the second draw.

The tree diagram showing the two draws and their probabilities is:

(i) P(red first, then blue)

= P(red on first draw) × P(blue on second draw, after a red is removed)

$= \dfrac{4}{9} \times \dfrac{5}{8} \\[1em] = \dfrac{20}{72} \\[1em] = \dfrac{5}{18}$

∴ Probability of drawing a red ball and then a blue ball = $\mathbf{\dfrac{5}{18}}$

(ii) P(2 blue balls)

= P(blue on first draw) × P(blue on second draw, after a blue is removed)

$= \dfrac{5}{9} \times \dfrac{4}{8} \\[1em] = \dfrac{20}{72} \\[1em] = \dfrac{5}{18}$

∴ Probability of drawing 2 blue balls = $\mathbf{\dfrac{5}{18}}$

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

When a pair of 6-sided dice is thrown, the sum of the two numbers can be any whole number from 2 (1 + 1) to 12 (6 + 6).

An event with probability 0 (impossible event): 'The sum of the two dice is 1.'

This can never happen, because the smallest possible sum is 2. So its probability is 0.

An event with probability 1 (certain event): 'The sum of the two dice is between 2 and 12 (both inclusive).'

This always happens, because every possible sum lies between 2 and 12. So its probability is 1.

Write the sample space and calculate the probability based on the given information.

(i) Two dice are rolled. What is the probability that the sum is a prime number greater than 5?

(ii) A bag contains 4 red, 3 green, and 2 blue balls. Two balls are drawn without replacement. What is the probability that both are of different colours?

(iii) Three coins are tossed. What is the probability that the first coin shows heads and exactly two heads occur in total?

(iv) A four-digit number is formed using the digits 1, 2, 3, and 4 with no repetition. What is the probability that the number is even?

(v) A student takes a multiple-choice test with 3 questions, each having 4 options (A, B, C, D), with only one correct answer. What is the probability that the student guesses and gets exactly 2 answers correct?

Answer

(i) When two dice are rolled, the total number of possible outcomes = 6 × 6 = 36.

The prime numbers greater than 5 that can occur as a sum (sums range from 2 to 12) are 7 and 11.

Ways to get a sum of 7 : (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 ways.

Ways to get a sum of 11 : (5,6), (6,5) → 2 ways.

Number of favourable outcomes = 6 + 2 = 8.

P (sum is a prime greater than 5) = $\dfrac{8}{36} = \dfrac{2}{9}$

∴ Required probability = $\mathbf{\dfrac{2}{9}}$

(ii) The bag has 4 red + 3 green + 2 blue = 9 balls.

Drawing 2 balls without replacement, total number of ways = $\dfrac{9 \times 8}{2}$ = 36.

When we pick the 1st ball we have 9 choices and after picking it only 8 balls are left, and since picking ball A then ball B is the same pair as ball B then ball A, we divide by 2.

It is easier to first find the number of ways both balls are of the same colour :

Both red = $\dfrac{4 \times 3}{2}$ = 6, Both green = $\dfrac{3 \times 2}{2}$ = 3, Both blue = $\dfrac{2 \times 1}{2}$ = 1.

We have 4 red balls so we pair any 2 of them, 3 green balls so we pair any 2 of them, and only 2 blue balls so there is just 1 pair possible.

Number of ways both are the same colour = 6 + 3 + 1 = 10.

Since counting different colour pairs directly is complicated, we subtract same colour pairs from total.

Number of ways both are of different colours = 36 - 10 = 26.

P (both different colours) = $\dfrac{26}{36}$ = $\dfrac{13}{18}$

∴ Required probability = $\mathbf{\dfrac{13}{18}}$

(iii) When three coins are tossed, the sample space is :

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}, so n(S) = 8.

We need the first coin to be a Head and exactly two heads in total.

The outcomes with the first coin H are : HHH, HHT, HTH, HTT.

Among these, the ones with exactly two heads are : HHT and HTH → 2 outcomes.

P (first coin head and exactly two heads) = $\dfrac{2}{8}$ = $\dfrac{1}{4}$

∴ Required probability = $\mathbf{\dfrac{1}{4}}$

(iv) Using the digits 1, 2, 3, 4 with no repetition, the total number of four-digit numbers = 4 × 3 × 2 × 1 = 24.

We have 4 choices for the 1st place, 3 remaining for the 2nd place, 2 remaining for the 3rd place and 1 remaining for the 4th place.

For the number to be even, the units digit must be 2 or 4 $\rightarrow$ 2 choices.

A number is even only when its last digit is even, and from 1, 2, 3, 4 the only even digits are 2 and 4.

Once we fix the units digit, 3 digits are left to fill the remaining 3 places = 3 × 2 × 1 = 6 ways.

Number of favourable (even) numbers = 2 × 6 = 12.

P (number is even) = $\dfrac{12}{24}$ = $\dfrac{1}{2}$

∴ Required probability = $\mathbf{\dfrac{1}{2}}$

(v) For each question, P (correct) = $\dfrac{1}{4}$ and P (wrong) = $\dfrac{3}{4}$.

Each question has 4 options and only 1 is correct, so the chance of guessing correctly is 1 out of 4, and the chance of getting it wrong is 3 out of 4.

The student must get exactly 2 of the 3 questions correct.

Number of ways to choose which 2 questions are correct = $\dfrac{3 \times 2}{2}$ = 3.

This means the 2 correct answers could be questions (1,2) or (1,3) or (2,3), giving us 3 possible combinations.

P (exactly 2 correct) = 3 × $\left(\dfrac{1}{4}\right)^2$ × $\dfrac{3}{4}$

Here $\left(\dfrac{1}{4}\right)^2$ is the probability of getting 2 questions correct, and $\dfrac{3}{4}$ is the probability of getting the remaining 1 question wrong.

$\text {P (exactly 2 correct)} = 3 \times \dfrac{1}{16} \times \dfrac{3}{4} \\[1em] = \dfrac{9}{64}$

∴ Required probability = $\mathbf{\dfrac{9}{64}}$

A box contains 4 balls numbered 1 to 4. Record a sample space using a tree diagram for the following experiments:

(i) A ball is drawn, and the number is recorded. Then the ball is returned, and a second ball is drawn and recorded.

(ii) A ball is drawn and recorded. Without replacing the first ball, the experimenter draws and records a second ball.

(iii) What are the sizes of these two sample spaces?

Answer

(i) With replacement: the first ball is returned, so the second draw can again be any of 1, 2, 3, 4.

The tree diagram for this experiment is:

Sample space:

S = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4), (4,1), (4,2), (4,3), (4,4)}

(ii) Without replacement: the first ball is not returned, so the second number must be different from the first.

The tree diagram for this experiment is:

Sample space:

S = {(1,2), (1,3), (1,4), (2,1), (2,3), (2,4), (3,1), (3,2), (3,4), (4,1), (4,2), (4,3)}

(iii) Size of the first sample space (with replacement) = 4 × 4 = 16.

Size of the second sample space (without replacement) = 4 × 3 = 12.

∴ The two sample spaces have sizes 16 and 12 respectively.

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

The coin can show Head (H) or Tail (T), and the card drawn can be numbered 1, 2, 3, 4, 5 or 6.

Pairing each coin result with each card number, the sample space is:

S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}

∴ n(S) = 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?

(i) {1, 2, 3}

(ii) {0, 1, 2}

(iii) {0, 1, 2, 3, 4}

(iv) {0, 1, 2, 3}

Answer

When three coins are tossed, the number of heads can be 0, 1, 2 or 3. A correct sample space must list every one of these and nothing extra.

∴ Option (iv) {0, 1, 2, 3} is the correct sample space.

The other lists fail because:

(i) {1, 2, 3} — leaves out 0 heads (the outcome TTT), so it misses a possible outcome.

(ii) {0, 1, 2} — leaves out 3 heads (the outcome HHH), so it misses a possible outcome.

(iii) {0, 1, 2, 3, 4} — includes 4 heads, which is impossible with only three coins, so it contains an outcome that can never occur.

Suppose you drop a dye at random on the rectangular region shown in Fig. 7.8. What is the probability that it will land inside the circle with a diameter of 1 m?

Answer

This is a geometric probability, found by comparing the area of the favourable region (the circle) with the area of the whole region (the rectangle).

The rectangle measures 3 m by 2 m.

Area of the rectangle = 3 × 2 = 6 m².

The circle has a diameter of 1 m, so its radius = $\dfrac{1}{2}$ m.

Area of the circle = $\pi r^2$ = $\pi \times \left(\dfrac{1}{2}\right)^2$ = $\dfrac{\pi}{4}$ m².

P (lands inside the circle) = $\dfrac{\text{area of the circle}}{\text{area of the rectangle}}$

$= \dfrac{\dfrac{\pi}{4}}{6} \\[1em] = \dfrac{\pi}{24}$

Taking $\pi = \dfrac{22}{7}$ :

$= \dfrac{22}{7 \times 24} \\[1em] = \dfrac{22}{168} \\[1em] = \dfrac{11}{84} \approx 0.131$

∴ Probability that the dye lands inside the circle = $\mathbf{\dfrac{\pi}{24}}$ ≈ 0.131