Chapter 7
The Mathematics of Maybe: Introduction to Probability
Exercise Set 7.1
1. 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.
Ans :
Ranking : 1
Label : Certain
Reason : In every week, Monday always comes after Sunday.
(ii) It will snow in Mumbai in July.
Ans :
Ranking : 0
Label : Imposible
Reason : Mumbai has a tropical climate, and it does not snow there in July.
(iii) An elephant will walk through your classroom today.
Ans :
Ranking : 0.1
Label : Less Likely
Reason : It is possible in very unusual situations, but it is extremely unlikely to happen.
(iv) You will greet at least one friend at school tomorrow
Ans :
Ranking : 0.9
Label : More Likely
Reason : If school is open and you attend, you are likely to meet and greet at least one friend.
Exercise Set 7.2
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.
Ans :
No. of red sweets = 10
No. of green sweets = 8
No. of yellow sweets = 7
No. of blue sweets = 5
Total sweets in the sample
= 10 + 8 + 7 + 5
= 30
(i) Probability of picking a green sweet
P(E) = Number of favourable outcomes / Total number of possible outcomes
P(Green Sweets)
= 8/30
= 4/15
(ii)
P(Yellow Sweets) = 7/30
No. of Yellow Sweets / Total No. of Sweets = 7/30
No. of Yellow Sweets / 600 = 7/30
No. of Yellow Sweets
= (7/30)×600
=7×20
=140
Hence, no.of yellow sweets in the large bag is 140.
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.
Ans :
Science Club = 14 students
Arts Club = 11 students
Sports Club = 9 students
Debate Club = 6 students
Total students in the sample
= 14 + 11 + 9 + 6
= 40
(i) Probability that a randomly chosen student prefers the Arts Club
P(E) = Number of favourable outcomes / Total number of possible outcomes
P(Arts Club) = 11/40
(ii) P(Sports Club) = 9/40
No. of students in sports club / Total No. of Students in the school = 9/40
No. of students in sports club / 800 = 9/40
No. of Yellow Sweets
= (9/40)×800
= 9×20
= 180
Hence, out of 800 students 180 students are likely to prefer the Sports Club.
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?
Ans : For this particular question, we have literally flipped a coin 20 times and we got head 11 times and tail 8 times. Your answer could be different.
(i) Head = 11 times
(ii) Tail = 8 times
(iii) P(E) = Number of times the event occurred / Total Number of trials
P(Head) = 11/20
(iv) If we toss the coin once more, the probability of getting tails will be = 1/2
This is because each toss of a fair coin is independent of the previous tosses.
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.
 |
| Fig. 7.5 Paper cup landing positions (left to right) — bottom, top and side |
We have performed this activity and got the following result.
No. of times the paper cup :
Lands on its bottom = 34
Lands upside down on its top = 26
Lands on its side = 40
Total No. of Trials = 100
P(E) = Number of times the event occurred / Total Number of trials
P(Lands on its bottom) = 34/100 = 0.34
P(Lands upside down on its top) = 26/100 = 0.26
P(Lands on its side) = 40/100 = 0.40
5. What is the probability of getting an even number when rolling a fair 6-sided die?
Ans : When a fair 6-sides die is rolled then sample space will be
{1, 2, 3, 4,5 ,6} = 6
Even No. of Dice = {2, 4, 6} = 3
P(E) = Number of favourable outcomes / Total number of possible outcomes
P(Even Number) = 3/6 = 1/2
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?
Ans : Given:
No. of rolls = 12
Number of times '3' appears = 3
(i) Experimental probability of rolling a '3'
P(E) = Number of times the event occurred / Total Number of trials
P(3) = 3/12
= 1/4
Experimental Probability is 1/4
(ii) Theoretical probability of rolling a '3'
When a die is rolled once
Sample space = {1, 2, 3, 4,5 ,6} = 6
No. of times 3 comes = {3} = 1
P(E) = Number of favourable outcomes / Total number of possible outcomes
P(3) = 1/6
Theoretical probability = 1/6
(iii) The experimental probability is based on actual results, while the theoretical probability is based on all equally likely outcomes. With only 12 rolls, the experimental probability may differ from the theoretical probability due to chance.
As the number of rolls increases to 60, 600, or 6000, the experimental probability is expected to get closer and closer to the theoretical probability of 1/6.
Remember : The law of large numbers, as the number of trials increases the experimental probability approaches to theoretical numbers.
Exercise Set 7.3
1. When a single 6-sided die is rolled, what is the total number of possible outcomes in the sample space?
Ans : When a 6-sides die is rolled then sample space is {1, 2, 3, 4, 5, 6} = 6
Clearly total number of possible outcomes would be 6.
2. For the following experiments write down the sample space S.
(i) Rolling a die and tossing a coin together.
Ans :
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)}
(ii) Choosing a random integer between – 5 and + 5.
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
Ans : S = {Green, Red}
Note : No same outcome would be listed one than once in sample space. Even though the box contains 5 green balls and 7 red balls, the outcomes are simply Green or Red. We do not write Green five times or Red seven times because repeated outcomes are not listed more than once in a sample space.
3. 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.
Ans :
S = {(Samosa, Chai), (Samosa, Lassi), (Pakora, Chai), (Pakora, Lassi), (Bhaji, Chai), (Bhaji, Lassi)}
(ii) List the event ‘Selecting Samosa as a snack.’
The event "Selecting Samosa as a snack" is
E = {(Samosa, Chai), (Samosa, Lassi)}
Exercise Set 7.4
1. 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.
Ans :
(ii) List the sample space.
Ans :
S = {(Apple, Banana), (Apple, Mango), (Orange, Banana), (Orange, Mango), (Orange2, Banana), (Orange2, Mango)}
(iii) What is the probability of picking one apple and one banana?
Ans :
No. of favourable outcomes = {(Apple, Banana)} = 1
Total Possible outcomes = {(Apple, Banana), (Apple, Mango), (Orange, Banana), (Orange, Mango), (Orange2, Banana), (Orange2, Mango)} = 6
P(E) = Number of favourable outcomes / Total number of possible outcomes
P(one apple and one banana) = 1/6
2. 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?
Ans :
End-of-Chapter Exercises
1. Fill in the blanks.
(i) The probability of an impossible event is _______.
Ans : 0
(ii) The set of all possible outcomes of a random experiment is called the __________.
Ans : sample space
(iii) The probability of an event that is certain to happen is _______.
Ans : 1
(iv) Tossing a fair coin has a probability of ______ for getting heads.
Ans : 1/2
2. 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).
Ans :
Note : Relative frequency is another name of experimental frequency.
Total No. of trials = 50
No. of times event occurs = 15
Relative frequency = 15/50 = 3/10 = 0.3
In a survey of 50 students, 15 students said they liked football. The number of students who like football is 15, and the relative frequency is 0.3
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.
Ans : Not equally likely.
(ii) Tossing a fair coin once.
Ans : Equally likely.
(iii) Rolling a fair 6-sided die.
Ans : Equally likely.
(iv) Choosing a marble randomly from a bag that contains 3 red marbles and 7 blue marbles.
Ans : Not equally likely.
(v) A baby is born. It is a boy or a girl.
Ans : Equally likely.
4. 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?
Ans : Sample space:
S = {HH, HT, TH, TT}
Event (at least one head):
E = {HH, HT, TH}
Probability = Number of favourable outcomes / Total number of outcomes
= 3/4
(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?
Ans : Sample space:
S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Event (drawing an even number):
E = {2, 4, 6, 8, 10}
Probability = 5/10
= 1/2
(iii) A die is rolled once. What is the probability of getting a number greater than 4?
Ans : Sample space:
S = {1, 2, 3, 4, 5, 6}
Event (number greater than 4):
E = {5, 6}
Probability = 2/6
= 1/3
(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?
Ans :
Sample space:
S = {Red, Blue, Green}
Since the balls are not equally many, use the total number of balls.
Total balls = 6
Favourable outcomes (not red) = 2 blue + 1 green = 3
Probability = 3/6
= 1/2
(v) Three coins are tossed simultaneously. What is the probability of getting exactly two heads?
Ans :
Sample space:
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Event (exactly two heads):
E = {HHT, HTH, THH}
Probability = 3/8
5. A bag has 3 candies: strawberry, lemon, and mint. One is picked at random. What is the probability of picking a strawberry candy?
Ans :
The possible outcomes are:
S = {Strawberry, Lemon, Mint}
Number of favourable outcomes = 1
Total number of possible outcomes = 3
Probability of picking a strawberry candy = 1/3
6. 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.
Ans :
Shirt - Pant - Combination
Red - Jeans - (Red, Jeans)
Red - Khakis - (Red, Khakis)
Red - Shorts - (Red, Shorts)
Blue - Jeans - (Blue, Jeans)
Blue - Khakis - (Blue, Khakis)
Blue - Shorts - (Blue, Shorts)
Hence, there are a total of 6 outfit combinations.
7. A tyre company records distances before replacement in 1000 cases
= 535/1000
