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**Solving SBT Lesson 2: Get acquainted with the probability of random events (C9 SBT Math 7 Horizon)**

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### Solve problem 1 page 85 SBT Math 7 Creative horizon episode 2 – CTST

Roll a balanced six-sided dice. Calculate the probability of the following events:

A: “A face with 2 dots appears”

B: “Appears face with a number of dots divisible by 4”

C: “Appears face with number of dots divisible by 7”

D: “Appears a face whose number is a divisor of 60

**Detailed instructions for solving Lesson 1**

**Solution method**

Determine what is a sure event, an unlikely event, and a random event to calculate the probability

**Detailed explanation**

Since the dice are symmetrical, its 6 faces have the same probability.

– Since there is only one side with 2 dots, \(P(A) = \frac{1}{6}\)

– Since there is only one face whose number of dots is divisible by 4, \(P(B) = \frac{1}{6}\)

– Since no face has a dot divisible by 7, C is an impossible event, so \(P(C) = 0\).

– Since all 6 faces have a dot that is a divisor of 60, the event D is a sure event, so \(P(D) = 1\)

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### Solution 2 page 85 SBT Math 7 Creative horizon episode 2 – CTST

On the wall there is a circular disk with uniform and balanced structure (Figure 3). The disc is divided into 12 equal fan shapes and is numbered from 1 to 12. Hoang rotates the disc around the axis mounted on the center and observes what number the arrow points to when stopped. Calculate the probability of the following events:

A: “The arrow points to box 7”

B: “The arrow points to the odd number box”

C: “The arrow points to the box where it is greater than 11”

**Detailed instructions for solving Lesson 2**

**Solution method**

Calculate the number of outcomes for each event

**Detailed explanation**

– Since the 12 fan shapes are equal, the probability that the arrow points to each fan shape is equal. Hence \(P(A) = \frac{1}{{12}}\).

– Since the fraction of odd-numbered fan shapes is the same size as even-numbered fan shapes, the probability of event B occurring is \(P(B) = \frac{1}{2}\).

– Since there is only one fan with a number greater than 11, the probability of event C is \(P(C) = \frac{1}{{12}}\)

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### Solution 3 page 85 SBT Math 7 Creative horizon episode 2 – CTST

A closed box contains 5 balls of the same size and weight with the numbers 5, 10, 15, 20, 25 respectively. Pick 1 ball at random from the box. Calculate the probability of the following events:

A: “The ball taken out is marked with a prime number”

B: “The ball is taken out and the number is divisible by 5”

C: “The ball is taken out and the number is divisible by 3”

D: “Pull out the number is a multiple of 6”

**Detailed instructions for solving Lesson 3**

**Solution method**

Calculate the number of outcomes for each event

**Detailed explanation**

– All 5 balls have the same size and mass; There is only 1 ball with the number 5 being a prime number so \(P(A) = \frac{1}{5}\).

– Event B is a sure event because all 5 numbers on each ball are divisible by 5. So

P(B) = 1.

– Only the ball with the number 15 ends in 3 of the 5 balls so \(P(C) = \frac{1}{5}\)

– There are no balls with numbers that are multiples of 6, so event D is an impossible event. So

\(P(C) = 0\)

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### Solve problem 4 page 86 SBT Math 7 Creative horizon episode 2 – CTST

A closed box contains 5 blue balls, 5 red balls and 5 white balls of the same size and mass. Randomly pick 1 ball from the box. Find the probability that the ball drawn is green.

**Detailed instructions for solving Lesson 4**

**Solution method**

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Calculate the number of outcomes for each event

**Detailed explanation**

Since the number of blue, red and white balls are equal and the bulbs have the same size and mass, all 3 colors have the same chance of being selected. Therefore, the probability that the selected event is green is \(\frac{1}{3}\)

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### Solve problem 5 page 86 SBT Math 7 Creative horizon episode 2 – CTST

In the box there are 1 blue marble, 1 white marble and 1 red marble of the same size and weight. Randomly draw 2 marbles from the box. Calculate the probability of the following event:

A: “Two marbles are drawn of the same color”

B: “None of the two selected marbles are blue or white.”

**Detailed instructions for solving Lesson 5**

**Solution method**

A and B are both impossible events, so the probability is 0 .

**Detailed explanation**

Variables with A and B are both impossible events because in the box there is only 1 blue marble, white marble and 1 red marble. So P(A) = 0; P(B) = 0

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### Solution 6 page 86 SBT Math 7 Creative horizon episode 2 – CTST

The chart below shows the rainfall (unit: mm) of Lai Chau and Ca Mau provinces in the years 2016 – 2020. Randomly select 1 year in those 6 years. Calculate the probabilities of the following events:

A: “In the selected year, the rainfall in Ca Mau is higher than in Lai Chau”

B: “In the selected year, the rainfall in Ca Mau is less than 25 m”

C: “In the selected year, the rainfall in Lai Chau is twice the amount in Ca Mau

**Detailed instructions for solving Lesson 6**

**Solution method**

Calculate the number of outcomes that occur for each event.

**Detailed explanation**

– There is only one year 2016 in 5 years from 2016 – 2020 the rainfall in Ca Mau is higher than in Lai Chau, so \(P(A) = \frac{1}{5}\)

– In the whole 5 years from 2016 to 2020, the rainfall in Ca Mau is less than 25 m, so event B is a certain event. So \(P(B) = 1\)

– There is no year in the 5 years from 2016 to 2020 that rainfall in Lai Chau is twice as much as in Ca Mau, so event C is an impossible event. So \(P(C) = 0\)

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### Solve problem 7 page 86 SBT Math 7 Creative horizon episode 2 – CTST

Toss two equal and equal coins. Compare the probabilities of the following events:

A: “There are no two tails”

B: “Both are heads”

C: “Have at least one tail”

**Detailed instructions for solving Lesson 7**

**Solution method**

Compare three events A, B, and C to see which event occurs more often

**Detailed explanation**

Event A always happens, so P(A) = 1

Since when B occurs, C also occurs, so the probability of C happening is higher than that of B. Therefore:

P(B) < P(C)

So P(B) < P(C) < P(A)

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### Solve problem 8 page 86 SBT Math 7 Creative horizon episode 2 – CTST

Cuong’s computer password consists of 8 characters, of which the first 2 are numbers, the last 6 are letters. Unfortunately, Cuong forgot the first character. Cuong chose 2 digits at random and tried to open the calculator. Calculate the probability that Cuong can open the calculator.

**Detailed instructions for solving Lesson 8**

**Solution method**

Calculate the probabilities for the first two characters

**Detailed explanation**

Since 00 to 99 have 100 numbers, there are 100 possibilities for the first 2 characters. The probability that Cuong can open the calculator is: \(\frac{1}{{100}}\)

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