Probability is the ideal of relative frequency – Multiple Choice Questions

In Class IX, you have studied about experimental (or empirical) probabilities of events which were based on the results of actual experiments. We discussed an experiment
of tossing a coin 1000 times in which the frequencies of the outcomes were as follows:

Head : 455 Tail: 545

Based on this experiment, the empirical probability of a head is 455/1000, i.e., 0.455 and that of getting a tail is 0.545. (Also see Example 1, Chapter 15 of Class IX Mathematics Textbook.) Note that these probabilities are based on the results of an actual experiment of tossing a coin 1000 times. For this reason, they are called experimental or empirical probabilities. In fact, experimental probabilities are based on the results of actual experiments and adequate recordings of the happening of the events. Moreover, these probabilities are only ‘estimates’. If we perform the same experiment for another 1000 times, we may get different data giving different probability estimates.

In Class IX, you tossed a coin many times and noted the number of times it turned up heads (or tails) (refer to Activities 1 and 2 of Chapter 15). You also noted that as the
number of tosses of the coin increased, the experimental probability of getting a head (or tail) came closer and closer to the number ½. Not only you, but many other persons from different parts of the world have done this kind of experiment and recorded the number of heads that turned up.

For example, the eighteenth century French naturalist Comte de Buffon tossed a coin 4040 times and got 2048 heads. The experimental probabilility of getting a head, in this case, was 2048/4040. i.e., 0.507. J.E. Kerrich, from Britain, recorded 5067 heads in 10000 tosses of a coin. The experimental probability of getting a head, in this case, was 5067/10000=0.5067. Statistician Karl Pearson spent some more time, making 24000 tosses of a coin. He got 12012 heads, and thus, the experimental probability of a head obtained by him was 0.5005.

Now, suppose we ask, ‘What will the experimental probability of a head be if the experiment is carried on upto, say, one million times? Or 10 million times? And so on?’
You would intuitively feel that as the number of tosses increases, the experimental probability of a head (or a tail) seems to be settling down around the number 0.5, i.e., ½, which is what we call the theoretical probability of getting a head (or getting a tail), as you will see in the next section. In this chapter, we provide an introduction to the theoretical (also called classical) probability of an event, and discuss simple problems based on this concept.

Relative frequency

Probability of sample points (that from the sample space) cannot always be considered the same. For example in tossing of a coin which is not balanced, then the probability of getting each side of the coin is not the same. Another example is the attempt to shoot an object, the probability of a shot hit the target is not equal to the probability of a shot does not hit target.

If each sample point can not be considered the same, then the probability of each sample point is determined based on the results of the experiment. By doing the experiment over and over again we can note the number of sample points. Probability of the sample point is the comparison of the number of occurrence of sample points with the number of experiments. The method to get the probability like this is known as the definition of probability basedon relative frequency.

Example 1:
Suppose a dice is thrown 30 times. And the emerging number is a recorded and the results are presented in the following table.

emerging dice frequency table

Determine the relative frequency (probability) of:
a. The emerging of dice’s eyes 3
b. The emerging of dice’s eyes 4
Answer:
The number of experiments =4+3+6+7+5+5=30
a. The emerging of dice’s eyes 3 is equal to 6.
Relative frequency (probability) of the emerging dice’s eyes 3=6/30
b. The emerging of dice’s eyes 3 is equal to 7.
Relative frequency (probability) of the emerging dice’s eyes 4=7/30

Example 2:
The eggs supplied by a poultry farm during a week broke during transit as follows:
1%,2%,1½%,½%,1%,2%,1%
Find the probability of the eggs that broke in a day. Calculate the number of eggs that will be broken in transiting the following number of eggs:
Solution: 1+2+1½+½+1+2+1=9
Number of eggs broken in a week =(9%)/7=9/700
(i). 7,000
Number of eggs broken 1 of 7,000 eggs =7000×9/700=90 eggs.
(ii). 8,400
Number of eggs broken 1 of 8,400 eggs =8000×9/700=108 eggs.
(iii). 10,500
Number of eggs broken 1 of 10,500 eggs =10500×9/700=135 eggs.

Ex3. Mr. Kadi, a textile seller recorded the number of textiles sold per day for 100 days. From these records are obtained the following information:

description table of sold textiles

Relative frequency of textiles which are sold in length of 30 m is equal to …
A. 0.1 B. 0.2 C. 0.3 D. 0.4 E. 0.5
Answer: D, explanation:
f1

Ex4. In tossing of a dice for 100 times, recorded results are as follows.
table of tossing dice

In these experiments, the relative frequency of the emerging of dice’s eye 6=⋯
A. 0.14 B. 0.15 C. 0.16 D. 1.5 E. 0.06
Answer: B, explanation:

f2

Ex5. If f states the value of relative frequency of a compound event, then the value of f is …
A. P>1 B. P<1 C. 0<P<1 D. 0≤P≤1 E. -1≤P≤1
Answer: C, explanation:
Clear!

Use the following passage to answer examples the number 6-8
City PQR has 1,500,000 citizens.
If all citizens are grouped according to age and sex, then the frequency table is as follows.

table of relative frequency and citizens

Ex6. The number of female in city PQR is equal to … people.
A. 900,000 B. =800,000 C. 700,000 D. 600,000 E. 500,000
Answer: A, explanation:

f=P×n
f=0.60×1500000=900000

Ex7. The number of citizens in age less than 22 years in city PQR is equal to … people.
A. 350,000 B. 365,000 C. 375,000 D. 450,000 E. 475,000
Answer: C, explanation:

f=(0.15+0,10)×1500000=375000

Ex8. The number of male citizens in age 22 until 45 years in city PQR is equal to … people.
A. 150,000 B. 200,000 C. 250,000 D. 300,000 E. 350,000
Answer: A, explanation:
f=0.1×1500000=150000

Ex9. A die is tossed 100 times. The result is tabulated below. Study the table and answer the questions given below the table:
Event Tally Marks Frequency

table of frequencies

(i). How many times do 3 times appear?
Ans: Dots 3 appears 20 times.
(ii). How many times do 5 dots appear?
Ans: Dots 5 appears 15 times.
(iii). How many times does an even number of dots appear?
Ans: Even dots are 2,4,6, which appears 17,18 and 16 times respectively, and total numbers are 17+18+16=51 times.
(iv). How many times does a prime number of dots appear?
Ans: Prime dots are 1,3,5, which appears 14,20 and 15 times respectively, and total numbers are 14+20+15=52 times.
(v). Find the probability of each one of the above cases?
Ans:
probability f

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