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**Solve exercises at the end of chapter 9 – Math 7 Horizons**

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### Solution 1 page 96 Math textbook 7 Creative horizon volume 2 – CTST

There are 3 comic books and 1 textbook on the bookshelf. An randomly selects 2 books from the bookshelf. Which of the following events is certain, unlikely, or random. Why?

A: “An chooses 2 comic books”.

B: “An chooses at least 1 comic book”.

C: “An chooses 2 textbooks”.

**Detailed instructions for solving Lesson 1**

**Solution method**

We rely on the correctness of the event and then consider it as a sure, impossible or random event

**Detailed explanation**

Event A is a random event because it is possible for An to choose 2 comic books but also 1 comic book and 1 textbook.

Event B is a sure event because An taking 2 books at random will always get 1 comic book

Event C is an impossible event because there is only one textbook on the shelf.

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### Solution 2 page 96 Math textbook 7 Creative horizon volume 2 – CTST

Roll two balanced dice. Compare the probabilities of the following events:

A: “The total number of dots appearing on the top of the two dice is even”,

B: “The number of dots appearing on the top two dice is 6”,

C: “The number of dots appearing on the top of two dice is equal”.

**Detailed instructions for solving Lesson 2**

**Solution method**

We calculate the probability of events A, B, C and then compare the events

**Detailed explanation**

Event C: the same number of dots on 2 faces will have an even result, but since event C requires 2 dice rolled with the same number of dots, event A only needs the sum of the dots on the 2 dice to be even So if 1 dice rolls 3 and 1 dice rolls out 1, event A still occurs. So P(C) < P(A)

Event B : the number of dots on both sides is 6 (total is 12 dots) with probability \(\frac{1}{{11}}\) and there is only 1 outcome so P(B) will be correct. lowest capacity.

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### Solution 3 page 96 Math textbook 7 Creative horizon volume 2 – CTST

A box has 4 cards of the same size and are numbered 2, 4, 6, 8 respectively. Pick 1 card at random from the box. Calculate the probabilities of the following events:

A: “Get a card with a prime number”

B: “Get an odd number card”

C: “Get an even number card

**Detailed instructions for solving Lesson 3**

**Solution method**

We consider the probabilities of the events and then compute the probabilities

**Detailed explanation**

Event A is a random event because 2 is a prime number.

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The number of possible outcomes is 4. The number of possibilities for event A is 1 (get card number 2)

\( \Rightarrow P(A)=\dfrac{1}{4}\)

Since none of the cards have an odd number, B is an impossible event, so P(B) = 0

Since all cards are even, C is a sure event \( \Rightarrow P(C) = 1\)

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### Solution 4 page 96 Math textbook 7 Creative horizon volume 2 – CTST

A closed box contains 5 balls of equal size and mass, of which 1 are blue and 4 are red. Pick a ball at random from the box, calculate the probability of the following events

A: “The ball taken out is yellow”

B: “The ball taken out is blue”.

**Detailed instructions for solving Lesson 4**

**Solution method**

We calculate the number of outcomes (n) and then calculate the probability of the events occurring.

**Detailed explanation**

Since there is no yellow ball in the box, P(A) = 0

The number of results is 1 + 4 = 5.

The number of chances that event B will occur is 1.

\( \Rightarrow P(B) = \dfrac{1}{5}\)

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### Solve problems 5 pages 96 Math textbook 7 Creative horizon volume 2 – CTST

The chart below shows the number of junior high school students in Phu Tho province from 2010 to 2019.

Randomly select a year in that period. Know the probability of choosing each year is the same

a) State the set of possible outcomes for the selected year.

b) Let B be the event: “Phu Tho province has over 85,000 junior high school students in the selected year”. Calculate the probability of the event.

**Detailed instructions for solving Lesson 5**

**Solution method**

a) We find all possible years and write down all possible outcomes

b) We calculate the number of possible outcomes, the number of possibilities of event B, and then find P(B)

**Detailed explanation**

a) Possible outcomes are: {2010;2011;2012;2013;2014;2015;2016;2017;2018;2019}

b) Number of possible outcomes n = 10.

The number of chances that event B will occur is: 1

\( \Rightarrow P(B) =\dfrac{1}{{10}}\)

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