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**Solve Exercises Lesson 2: Get acquainted with the probability of random events (C9 Math 7 Horizons)**

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### Solution 1 page 93 Math textbook 7 Creative horizon volume 2 – CTST

A circular card is divided into 6 equal parts as shown in Figure 1. Minh puts the card upright on the table, turns the arrow in the center and observes which box the arrow points to when he stops.

Compare the probabilities of the following events:

A: “The arrow points to the red box”

B: “The arrow points to the box with the number 3”

C: “The arrow points to the box where the number is greater than 2”

**Detailed instructions for solving Lesson 1**

**Solution method**

Probability of event A = Number of possibilities A : Total number of possibilities

**Detailed explanation**

In the stele, we see that 2 out of 6 cells are red, so the probability of returning to the red square is \(\frac{2}{6} = \frac{1}{3}\)

In the stele, we see that there is only 1 box 3, so the probability of returning to square 3 is \(\frac{1}{6}\)

In 6 cells, we see that there are 4 cells greater than 2, so the probability of returning to the box with a number greater than 2 is \(\frac{4}{6} = \frac{2}{3}\)

So the probability of event B is the lowest and the probability of event C is the highest

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### Solution 2 page 93 Math textbook 7 Creative horizon volume 2 – CTST

A box contains 100 cards of the same type, of which only one is marked as Lucky Card. Vase draws a card at random. Calculate the probability that the card drawn is a lucky card.

**Detailed instructions for solving Lesson 2**

**Solution method**

We will calculate the probability of taking 1 out of 10 cards from there to see whose probability is higher.

**Detailed explanation**

The probability of Binh taking 1 card is \(\frac{1}{{100}}\)

Minh’s probability of taking 10 cards is \(\frac{{10}}{{100}} = \frac{1}{{10}}\)

Since \(\frac{1}{{100}} < \frac{1}{{10}}\) the probability of getting Minh's lucky card is higher than Binh's.

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### Solve problem 3 page 94 Math textbook 7 Creative horizon volume 2 – CTST

Roll a balanced dice. Find the probabilities of the following events

a) A: “Sowing the face with the number of dots equal to 4”

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b) B: “Grow the face with the number of dots divisible by 5”

c) C: “Sowing the face with the number of dots is round ten”

**Detailed instructions for solving Lesson 3**

**Solution method**

We consider the probabilities of the events and then compare those probabilities.

**Detailed explanation**

a) Event A: Since there are 4 dots on a dice with 1 face, the probability of rolling a 4-dot face is \(\dfrac{1}{6}\)

b) Event B : since only 5 dots on faces are divisible by 5, the probability of rolling 5 dots is \(\dfrac{1}{6}\)

c) Event C: Since the number of dots on each side of the dice is between 1 and 6, event C is an impossible event. Therefore, the probability of event C occurring is 0.

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### Solution 4 page 94 Math textbook 7 Creative horizon volume 2 – CTST

The dance team has 1 male and 5 female friends. Randomly select 1 person to interview. Know that each of you is equally likely to be selected. Calculate the probability of the event that you are selected to be a male.

**Detailed instructions for solving Lesson 4**

**Solution method**

We calculate the ratio between male and female friends and then calculate the probability to choose a male friend

**Detailed explanation**

Since you have 1 boyfriend out of 5, the probability of you being selected as a male is \(\dfrac{1}{{1 + 5}} = \dfrac{1}{6}\)

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### Solve problem 5 page 94 Math textbook 7 Creative horizon volume 2 – CTST

The amount of electricity consumed per day in the first 5 days of September 2021 of a household is shown in the following chart. Randomly pick one day out of those 5 days. Calculate the probability of the event: “Household uses 10 kWh of electricity in the selected day”

**Detailed instructions for solving Lesson 5**

**Solution method**

We find all the days with 10kWh electricity consumption and then calculate the probability that the day has 10kWh electricity consumption in the first 5 days of September 2021.

**Detailed explanation**

In the first 5 days of September 2021, based on the chart, we can see that there is only one day on September 3 with a power consumption of 10kWh. Should randomly choose 1 day of electricity consumption of 10kWh, the probability of choosing is \(\dfrac{1}{5}\)

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