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.
Solution:
(i) Number of green sweets = 8
Total number of sweets = 30
Probability of picking a green sweets = \frac{\text{Number of green sweets}}{\text{Total number of sweets}}
= \frac{8}{30}=\frac{4}{15}=0.2666…\approx 0.267
(ii) Number of yellow sweets = 7
Total number of sweets = 30
Probability of picking a yellow sweets = \frac{\text{Number of green sweets}}{\text{Total number of sweets}}
= \frac{7}{30}
So, the estimated number of yellow sweets in 600 sweets = \frac{7}{30}×600
= 7×20
= 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.
Solution:
(i) Number of students who prefer arts club = 11
Total number of students in sample = 40
P(Arts Club) = \frac{\text{Number of students who prefer arts club}}{\text{Total number of students in sample}}
= \frac{11}{40}\approx 0.275
(ii) Number of students who prefer sports club = 9
Total number of students in sample = 40
P(Sports Club) = \frac{\text{Number of students who prefer sports club}}{\text{Total number of students in sample}}
= \frac{9}{40}
Estimated number of students in the whole school who prefer sports club = \frac{9}{40}×800
= 9×20
= 180
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?
Solution:
Suppose the coin was tossed 20 times and the results were:
Heads (H): 12 times
Tails (T): 8 times {Your answer may vary when you toss a coin 20 times}
(i) 12 times
(ii) 8 times
(iii) Experimental probability of getting heads = \frac{\text{Number of heads}}{\text{Total number of tosses}}
=\frac{12}{20}=\frac{3}{5}=0.6
(iv) Probability of getting tails \frac{1}{2}
Each toss of a fair coin is independent. The previous results do not affect the outcome of the next toss. A fair coin has two equally likely outcomes. Therefore, the probability of getting tails remains \frac{1}{2}
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.
Solution:
Suppose the paper cup is tossed 100 times and the results are:
| Outcome | Number of Times |
|---|---|
| Lands on its bottom | 42 |
| Lands upside down on its top | 18 |
| Lands on its side | 40 |
| Total | 100 |
The experimental probability of an event is:
Experimental Probability =\frac{\text{Number of times the event occurs}}{\text{Total number of trials}}
(i) Probability of landing on its bottom
P(\text{Bottom})=\frac{42}{100}=0.42(ii) Probability of landing upside down on its top
P(\text{Top})=\frac{18}{100}=0.18(iii) Probability of landing on its side
P(\text{Top})=\frac{40}{100}=0.405. What is the probability of getting an even number when rolling a fair 6-sided die?
Solution:
A fair 6-sided die has the numbers: {1,2,3,4,5,6}
The even numbers are: {2,4,6}
So,
Number of favourable outcomes = 3
Total number of possible outcomes = 6
The probability of getting an even number is:
P(\text{Even number})=\frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}
= \frac{3}{6}=\frac{1}{2}=0.5
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?
Solution:
(i) Total number of trials = 12
Number of times ‘3’ occurs = 3
Experimental probability of rolling a ‘3’
P(rolling a ‘3’) = \frac{\text{Number of times ’3’ occurs}}{\text{Total number of trails}}
= \frac{3}{12}=\frac{1}{4}
(ii) A fair 6-sided die has six equally likely outcomes: {1,2,3,4,5,6}
Theoretical probability of rolling a ‘3’
P(3) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}
=\frac{1}{6}
(iii) Since the die was rolled only 12 times, the experimental probability may differ from the theoretical probability due to chance.
If the die is rolled 60, 600, or 6000 times, the experimental probability is expected to get closer and closer to the theoretical probability of \frac{1}{6} because a larger number of trials gives more reliable results.
