**Expected Frequency**

If experiments are done over and over, then we can expect how many times an event can occur. For example, in a coin tossing for 10 times,we can expect the emergence of numeral side as much as 5 times. The expected of the number of the event which is appeared (or successful) in experiments that are done over and over is called expected frequency.

If an experiment is done as much as *N* times, and the probability of event K →*P*(K) then the expected frequency of the emerging of event K →*P*(K)∙*N*.

Example 8:

Suppose a balance dice is tossed for 30 times. Determine the expected frequency of the emerging of dice’s eyes is 3.

Answer:

In the tossing of a balance dice, the probability of the emerging of dice’s eyesis 3 →⅙. The number of experiments =30. Expected frequency of the emerging of number dice’s eyes is 3 →⅙×30=5.

**Exercise Competency test 7**

1. A balance dice is thrown for 90 times. The expected frequency of the emerging of dice’s eyes is 4 →⋯

A. 10 B. 12 C. 15 D. 20 E. 30

Answer: C, exposition:

*P*=⅙×90=15

A. 10 B. 15 C. 25 D. 30 E. 45

Answer: D, exposition:

Dice sides {3,6} are divisible by 3.

*P*=⅓×90=30

A. 2 B. 3 C. 4 D. 5 E. 6

Answer: C, exposition:

The set of bridge card contains cards within 4 types. There are 1×4=4 Aces.

Probability of drawing two aces at once is

A. 70 B. 60 C. 40 D. 30 E. 20

Answer: A, exposition:

We need to count probability of none image side (E) facing up first, it means all numeral sides (N) in face up position.

Plenty elements of the sample space is

The probability of at least one image side in face up position is

The expected frequency is

*f*=⅞×80=70

If in one day, the number of vehicles which park =150, so the number of car is equal to …

A. 60 B. 70 C. 80 D. 90 E. 100

Answer: D, exposition:

*f=P×n*

*f*=0,6×150=90

6. In one box there are 6 good eggs and 2 ratten eggs. Two eggs are taken one by one randomly without replacement. If the taking is done for 112 times, then the expected frequency of the taken eggs are both good =…

A. 63 B. 60 C. 56 D. 53 E. 50

Answer: B, exposition:

7. In a multiple choice questions test, there were 100 questions, each with 4 options, a, b, c, d. Out of these 4 options, there was only one correct option, while the other options were wrong. Each student is awarded 4 marks for each correct answer and is penalized with 1 mark for each wrong answer. One student marked all the answers as ‘c’. What is the most probable score that this student should get?

a. 0 b. 25 c. -25 d. 100

correct: b, solution:

There are 4 options {a,b,c,d}. Therefore, in 100 questions, 25 questions can have ‘c’ as the correct option. So he would have got =25×4-75=25 marks.

8. If five dice are rolled then what is the most probable sum of all the outcomes of these dice?

a. 17 b. 18

c. Neither 17 nor 18 d. Either 17 or 18

correct: d, solution:

There is equally likely of getting any number from 1 to 6 on all the dice. So the average of all the possible outcome is the most probable sum.

So the outcomes will have five 1’s, five 2’s, five 3’s, five 4’s, five 5’s and five 6’s. Average =105/6=17.5

Hence probable sum can be either 17 or 18.

For example, here is sample space of sum of rolling two dice. Look at shaded area what surrounds the center.