 ## SBT Prize at the end of chapter 9 – SBT Math 7 Horizon – Math Book

SBT Prize at the end of chapter 9 – SBT Math 7 Horizon
=======

### Solution 1 page 87 SBT Math 7 Creative horizon episode 2 – CTST

A box contains 4 blue pens and 1 black pen. Manh randomly selects 2 pens from the box and finds two pens of the same color. Which of the following events occurs and which does not:

A: “The two pens taken out are blue”

B: “the two pens taken out are red”

C: “There is at least 1 red pen out of the two pens taken out”

Detailed instructions for solving Lesson 87

Solution method

See which events happen and which do not.

Detailed explanation

Event A occurs because there are 4 blue pens and 1 black pen, and 2 pens are chosen at random.

Event B does not occur because there is no red pen in the box.

Event C does not occur because there is no red pen in the box.

–>

— *****

### Solve problem 2 page 87 SBT Math 7 Creative horizon episode 2 – CTST

Group 3 has 6 friends Ha, Hien, Hiep, Huong, Hung and Khanh. Randomly select 1 person in the group. State the set of outcomes that make each of the following events happen:

A: “Your chosen name starts with the letter H”

B: “Is your name chosen, not the letter “g””

C: “Your chosen name contains an accent mark”

What is the outcome that causes all three of the above events to occur?

Detailed instructions for solving Lesson 2

Solution method

Write the set by listing the items that occur for each event.

Detailed explanation

A = {Ha ; Gentle; Hiep; Smell; Hung}

B = {Ha; Gentle; Hiep; Independence}

C = {Ha ; Gentle; Hung}

The results Ha and Hien cause all three of the above events to occur.

–>

— *****

### Solve problem 3 page 87 SBT Math 7 Creative horizon episode 2 – CTST

A box has 4 lots numbered 3; 5; 7; 9. Randomly draw 2 lots. Of the following events, indicate which is certain, impossible, random. Why?

A: “The sum of the numbers on the two lots is 11”

B: “The product of the numbers on the two lots is odd”

C: “The number on one lot is equal to the square of the number on the other lot”

Detailed instructions for solving Lesson 3

Solution method

Explain that event A is impossible, B is certain, and C is random

Detailed explanation

A is an impossible event because the sum of the numbers on the two lots must be even.

B is the event that it is certain that the numbers recorded on the lot are odd, so the product of both numbers is odd.

C is a random event because it occurs when the numbers 3 and 9 are drawn.

–>

— *****

### Solve problem 4 page 87 SBT Math 7 Creative horizon episode 2 – CTST

At first, Huong had 2 bills of 5000 dong and 3 bills of 10,000 dong. Huong dropped 2 bills. Which of the following events is certain, unlikely, or random. Why?

A: “The amount Huong dropped is 30,000 VND”

B: “The amount of Huong’s whip is 10,000 VND”

C: “The remaining incense is at least 20,000 VND”

Detailed instructions for solving Lesson 4

Solution method

Explain that event A is an unlikely event, B is a random event, and C is a certain event

Detailed explanation

– A is an impossible event because the total amount dropped does not exceed 20 00 VND.

– B is a random event because it happens when Huong drops 2 bills of 5000 dong.

– C is a sure event because if two bills are dropped with the highest denomination of 10 000 VND, the remaining amount is 20 000 VND.

–>

— *****

### Solve problem 5 page 87 SBT Math 7 Creative horizon episode 2 – CTST

A business randomly chooses 1 month in 2022 to conduct a customer gratitude promotion. Find the probability that the firm chooses a month with less than 30 days, given that all months have the same probability of being selected.

Detailed instructions for solving Lesson 5

Solution method

Only 1 month is less than 30 days from which to deduce the probability to find.

Detailed explanation

Only February has less than 30 days, so the probability that the business chooses a month with less than 30 days is: $$\frac{1}{{12}}$$

–>

— *****

### Solve problem 6 page 87 SBT Math 7 Creative horizon episode 2 – CTST

Uncle Luan randomly draws 1 card from a deck of 52 cards.

a) Calculate the probability of the event: “Uncle Luan draws an Ace card”.

b) Calculate the probability of the event: “Uncle Luan draws a red card”.

Detailed instructions for solving Lesson 6

Solution method

See how many outcomes for each event

Detailed explanation

a) There is only 1 Ace card out of 52 cards, so the probability of the event: “Uncle Luan draws an Ace card” is $$\frac{1}{{52}}$$.

b) Since there are 26 red cards and 26 black cards, the probability of drawing a red card is equal to the ability to draw a black card. So the probability of drawing a red card is: $$\frac{1}{2}$$

–>

— *****

### Solve problem 7 page 88 SBT Math 7 Creative horizon episode 2 – CTST

He called his mother but forgot the rightmost digit of the phone number. Just randomly pick a number for that last digit and make a call.

a) Calculate the probability that Chinh correctly called his mother’s number.

b) How many times must she call at least to be sure to get the correct phone number for her mother?

Detailed instructions for solving Lesson 7

Solution method

See how many outcomes for each event

Detailed explanation

a) Since there are 10 different rightmost digits, the probability that Chinh correctly calls his mother’s phone number is $$\frac{1}{{10}}$$.

b) She has to call at least 9 times to make sure she has the correct phone number for her mother.

–>

— *****

### Solve problem 8 page 88 SBT Math 7 Creative horizon episode 2 – CTST

The houses in An’s block have even numbers, from number 26 to number 84, respectively. Uncle Phuc randomly chooses 1 house in An’s street to wish Tet. Find the probability of the event that house An is chosen.

Detailed instructions for solving Lesson 8

Solution method

See how many outcomes for each event

Detailed explanation

Since there are 30 even numbers from 26 to 84, An’s street has 30 houses. The probability that Uncle Phuc can choose An is: $$\frac{1}{{30}}$$

–>

— *****

### Solve problem 9 page 88 SBT Math 7 Creative horizon episode 2 – CTST

A box contains 10 marbles of the same size and weight, of which 1 blue, 3 red and 6 white. Randomly draw 1 marble from the box. Compare the probabilities of the following events:

A: “The marble taken out is blue”;

B: “The marble taken out is red”;

C: “The marble taken out is white”;

D: “The marble taken out is purple”;

Detailed instructions for solving Lesson 9

Solution method

Compare the probability of each event A, B, C, D

Detailed explanation

Since no marbles are purple, P(D)=0

Since marbles with the same probability are selected, the number of blue marbles is less than the number of red marbles, and the number of red marbles is less than the number of yellow marbles, so 0 < P(A) < P(B) < P(C)

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

–>

— *****

### Solution 10 page 88 SBT Math 7 Creative horizon episode 2 – CTST

The selling prices of 4 types of stocks A, B, C, D at the end of December 31, 2020 and 2021 are shown in the following chart. Ms. Thuy randomly chooses to buy one of the above four types of stocks on June 1, 2021.

Calculate the probability of the events after comparing the two time points:

A: “Selected shares have reduced selling price”;

B: “Selected shares have a selling price of more than 5,000 VND”;

C: “Selected shares whose selling price increased by more than 25%”;

Detailed instructions for solving Lesson 10

Solution method

See which events are impossible and which are certain events to calculate the probability of each event.

Detailed explanation

Share A has an increased selling price: $$41025 – 34570 = 6455$$ and an increase $$\frac{{6455}}{{34570}}.100\% = 18.7\%$$

Stock B’s selling price increased: $$5770 – 5670 = 1000$$ and increased $$\frac{{1000}}{{5670}}.100\% = 1.76\%$$

Shares with increased selling price: $$35102 – 34565 = 537$$ and increase $$\frac{{537}}{{34565}}.100\% = 1.55\%$$

Share D’s selling price decreased by: $$12980 – 12345 = 668$$ and decreased by $$\frac{{668}}{{12983}}.100\% = 5.15\%$$

– Only stock D has a decrease in selling price out of 4 sold shares, so $$P(A) = \frac{1}{4}$$

– Only stock A has a selling price of more than 5,000 dong, so $$P(B) = \frac{1}{4}$$

– None of the selected stocks have a selling price increase of more than 25%, so event C is an impossible event. So $$P(C) = 0$$.

–>

— *****