Question 1

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.

Accepted Answer

(i) Total sweets in sample = 30
Number of green sweets = 8
Probability of picking a green sweet = Number of green sweets / Total sweets = 8/30 = 4/15

(ii) Total sweets in large bag = 600
Sample proportion of yellow sweets = 7/30
Estimated number of yellow sweets in large bag = (7/30) × 600 = 140 sweets — To find the probability of picking a green sweet, divide the number of green sweets by total sweets in the sample.
To estimate the number of yellow sweets in the large bag, use the proportion of yellow sweets in the sample and multiply by total sweets in the bag.

Question 2

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.

Accepted Answer

(i) Total students in sample = 40
Number who prefer Arts Club = 11
Probability = 11/40

(ii) Total students in school = 800
Proportion who prefer Sports Club = 9/40
Estimated number preferring Sports Club = (9/40) × 800 = 180 students — Probability is calculated by dividing the number of students preferring Arts Club by total sample size.
To estimate the number in the whole school, multiply the proportion from the sample by total students.

Question 3

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?

Accepted Answer

(i) Number of heads = (count from your experiment)
(ii) Number of tails = 20 - number of heads
(iii) Experimental probability of heads = (Number of heads) / 20
(iv) Probability of tails in a single toss = 1/2 (theoretical probability, since coin is fair) — Count the number of heads and tails from your 20 tosses.
Calculate experimental probability by dividing number of heads by total tosses.
Theoretical probability of tails in a fair coin toss is always 1/2.

Question 4

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.

Accepted Answer

Record the number of times the cup lands on bottom, top, and side out of 100 tosses.
Calculate experimental probability for each outcome as:
Probability = (Number of times outcome occurs) / 100
Example: If bottom occurs 60 times, P(bottom) = 60/100 = 0.6 — Experimental probability is found by dividing the frequency of each outcome by the total number of trials (100).
Use the recorded data to calculate probabilities for bottom, top, and side.

Question 5

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

Accepted Answer

Sample space for a 6-sided die: {1, 2, 3, 4, 5, 6}
Even numbers in sample space: {2, 4, 6}
Number of even numbers = 3
Total number of outcomes = 6
Probability of even number = 3/6 = 1/2 — Count the favorable outcomes (even numbers) and divide by total possible outcomes (6).

Question 6

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?

Accepted Answer

(i) Experimental probability = Number of times '3' occurred / Total rolls = 3/12 = 1/4

(ii) Theoretical probability of rolling a '3' on a fair 6-sided die = 1/6

(iii) The experimental probability differs from theoretical because of the small number of trials and randomness.
As the number of rolls increases (60, 600, 6000), the experimental probability is expected to get closer to the theoretical probability (1/6) due to the Law of Large Numbers. — Experimental probability is based on actual outcomes observed.
Theoretical probability is based on equally likely outcomes.
Differences occur due to chance in small samples.
Increasing trials reduces this difference.

Question 7

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

Accepted Answer

When a single 6-sided die is rolled, the possible outcomes are the numbers on the faces of the die: 1, 2, 3, 4, 5, and 6. Therefore, the total number of possible outcomes in the sample space is 6. — A standard die has 6 faces, each with a distinct number from 1 to 6. Each face represents a unique outcome when the die is rolled. Hence, the sample space consists of these 6 outcomes.

Question 8

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.

Accepted Answer

(i) Rolling a die and tossing a coin together:
Sample space S consists of all possible pairs where the first element is the outcome of the die (1 to 6) and the second element is the outcome of the coin (Heads (H) or Tails (T)).

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

(ii) Choosing a random integer between -5 and +5:
The integers from -5 to +5 inclusive are: -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5.

So, sample space S = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}

(iii) A box containing 5 green and 7 red balls. One ball is drawn at random:
The sample space consists of the colors of the ball drawn.

S = {Green, Red} — (i) Since the die has 6 outcomes and the coin has 2 outcomes, the combined sample space is the Cartesian product of these sets.
(ii) All integers from -5 to +5 inclusive are possible outcomes.
(iii) Since the balls are only green or red, the sample space is the set of these two colors.

Question 9

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.'

Accepted Answer

(i) The sample space consists of all possible combinations of one snack and one drink.
Snacks = {Samosa, Pakora, Bhaji}
Drinks = {Chai, Lassi}

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

(ii) Event 'Selecting Samosa as a snack' includes all outcomes where the snack is Samosa, regardless of the drink chosen.

Event E = {(Samosa, Chai), (Samosa, Lassi)} — Each snack can be paired with each drink, so the sample space is the Cartesian product of the two sets. The event of selecting Samosa includes all pairs where the snack is Samosa.

Question 10

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?

Accepted Answer

Solution:
(i) Tree diagram:
From Basket A: Apple (A), Orange (O1), Orange (O2)
From Basket B: Banana (B), Mango (M)
Possible pairs:
- A-B
- A-M
- O1-B
- O1-M
- O2-B
- O2-M

(ii) Sample space S = { (Apple, Banana), (Apple, Mango), (Orange, Banana), (Orange, Mango) }
Note: Since Basket A has two oranges, but they are identical fruits, the sample space can be simplified as:
S = { (A, B), (A, M), (O, B), (O, M) }

(iii) Probability of picking one apple and one banana:
Number of favorable outcomes = 1 (Apple from A and Banana from B)
Total number of outcomes = 3 fruits in A × 2 fruits in B = 3 × 2 = 6
Therefore, P = 1/6 — Step-by-step:
- Identify fruits in each basket.
- Draw tree diagram with branches for each fruit in Basket A and Basket B.
- List all possible pairs (sample space).
- Count total outcomes = 3 × 2 = 6.
- Count favorable outcomes for (Apple, Banana) = 1.
- Calculate probability = favorable / total = 1/6.

Question 11

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?

Accepted Answer

Solution:
Total pens = 3 (Red) + 4 (Black) + 2 (Green) = 9 pens.

(i) Possible outcomes for each pick: Red (R), Black (B), Green (G).
Since the pen is replaced after the first pick, the second pick has the same possibilities.
Tree diagram branches:
- First pick: R, B, G
- Second pick (for each first pick): R, B, G
Sample space consists of 3 × 3 = 9 outcomes:
(R,R), (R,B), (R,G), (B,R), (B,B), (B,G), (G,R), (G,B), (G,G)

(ii) Probability that both picks are of the same colour:
Calculate probability for each colour:
P(R) = 3/9 = 1/3
P(B) = 4/9
P(G) = 2/9
Probability both pick Red = P(R) × P(R) = (1/3) × (1/3) = 1/9
Probability both pick Black = (4/9) × (4/9) = 16/81
Probability both pick Green = (2/9) × (2/9) = 4/81
Total probability both same colour = 1/9 + 16/81 + 4/81 = (9/81) + (16/81) + (4/81) = 29/81 — Step-by-step:
- Calculate total pens.
- Identify possible outcomes for each pick.
- Draw tree diagram with first pick branches and second pick branches.
- Calculate individual probabilities for each colour.
- Multiply probabilities for both picks being the same colour.
- Sum probabilities for all colours to get total probability.

Question 12

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.

Accepted 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 1/2 for getting heads. — Probability of impossible event = 0 because it cannot occur.
Sample space is the set of all possible outcomes.
Probability of certain event = 1 because it will definitely occur.
For a fair coin, two equally likely outcomes (head or tail), so probability of head = 1/2.

Question 13

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).

Accepted Answer

The number of students who like football is 15, and the relative frequency is 15/50 = 0.3. — Relative frequency = (Number of students who like football) / (Total number of students) = 15/50 = 0.3.

Question 14

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.

Accepted Answer

(i) Not equally likely because the car may start or not start but the probabilities depend on the car's condition.

(ii) Equally likely because a fair coin has two outcomes with equal chance.

(iii) Equally likely because a fair die has six faces with equal chance.

(iv) Not equally likely because there are 3 red and 7 blue marbles, so probabilities differ.

(v) Not necessarily equally likely as the probability of boy or girl may not be exactly equal. — Equally likely outcomes mean all outcomes have the same chance of occurring.
(ii) and (iii) satisfy this condition.
(i), (iv), and (v) do not because probabilities differ or depend on external factors.

Question 15

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?

Accepted Answer

(i) Sample space = {HH, HT, TH, TT}
Number of outcomes with at least one head = 3 (HH, HT, TH)
Probability = 3/4

(ii) Sample space = {1,2,3,4,5,6,7,8,9,10}
Number of even numbers = 5 (2,4,6,8,10)
Probability = 5/10 = 1/2

(iii) Sample space = {1,2,3,4,5,6}
Numbers greater than 4 = {5,6}
Probability = 2/6 = 1/3

(iv) Total balls = 3+2+1=6
Not red balls = 2+1=3
Probability = 3/6 = 1/2

(v) Sample space size = 2^3 = 8
Outcomes with exactly two heads = {HHT, HTH, THH} = 3
Probability = 3/8 — For each part, list all possible outcomes (sample space), count favorable outcomes, then calculate probability as favorable/total outcomes.

The Mathematics of Maybe

The Mathematics of Maybe — Study Notes

Introduction

Experiments and Events

Theoretical Approach to Probability

Practice Questions — The Mathematics of Maybe

All 8 Chapters in Ganita Manjari (English)