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Solving exercises at the end of chapter 9 (Math 10 – Connection textbook) KNTT
——————

A. Quiz:

Lesson 9.13: A box has four types: blue, red, white and yellow. Randomly pick one out. Let E be the event: “Red marble is obtained”. The opposite event of E is the event

A. Get the blue pill.

B. Get a yellow or white tablet

C. Get the white tablet.

D. Get a yellow or white or green pill.

Solution method – See details

E and \(\overline E \) are two events for if and only if \(E \cup \overline E = \Omega \) and \(E \cap \overline E = \emptyset \)

Detailed explanation

Choose D.
Lesson 9.14: Draw a card at random from a box of 30 cards numbered from 1 to 30. The probability that the number on the drawn card is divisible by 5 is:

A. \(\frac{1}{{30}}\)

B. \(\frac{1}{5}\)

C. \(\frac{1}{3}\)

D. \(\frac{2}{5}\)

Solution method – See details

Numbers divisible by 5 are numbers ending in 0 or 5.

Detailed explanation

The number of elements of the sample space is \(n\left( \Omega \right) = 30\).

Let E be the event: “The number on the card drawn is divisible by 5”

We have \(E = \left\{ {5;10;15;20;25;30} \right\} \Rightarrow n\left( E \right) = 6\)

So the probability of event E is \(P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( \Omega \right)}} = \frac{ 1}{5}\).

Select REMOVE
Lesson 9.15: Roll two balanced dice. The probability that the total number of dots appearing on the two dice is not greater than 4 is:

A. \(\frac{1}{7}\)

B. \(\frac{1}{6}\)

C. \(\frac{1}{8}\)

D. \(\frac{2}{9}\)

Solution method – See details

Use the classical probability formula \(P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( \Omega \right)}}\).

Detailed explanation

The number of elements of the sample space is \(n\left( \Omega \right) = 36\)

Let E be the event \(E = \left\{ {\left( {1,1} \right);\left( {1;2} \right);\left( {1,3} \right); \left( {2 & ;1} \right);\left( {2;2} \right);\left( {3,1} \right)} \right\}\) deduce \(n\left ( E \right) = 6\)

So \(P\left( E \right) = \frac{6}{{36}} = \frac{1}{6}\).

Select REMOVE
Lesson 9.16: A group in class 10T has 4 girls and 3 boys. The teacher randomly selects two students from that group to join the class newspaper team. What is the probability that two people selected have one male and one female friend?

A. \(\frac{4}{7}\)

B. \(\frac{2}{7}\)

C. \(\frac{1}{6}\)

D. \(\frac{2}{{21}}\)

Solution method – See details

Use the classical probability formula \(P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( \Omega \right)}}\).

Detailed explanation

The way to choose 2 friends from 10 friends is \(C_{10}^2 \Rightarrow n\left( \Omega \right) = C_{10}^2 = 21\)

Let A be the event: “Two chosen friends have one boy and one girl”.

How to choose a male friend is: 3 ways to choose

How to choose a female friend is: 4 ways to choose

According to the multiplication rule we have \(n\left( A \right) = 3.4 = 12\)

So the probability of event A is \(P\left( A \right) = \frac{{n\left( A \right)}}{{n\left( \Omega \right)}} = \frac{ {12}}{{21}} = \frac{4}{7}\).

Choose A

B. Essay:

Solve problem 9.17 page 88 Math textbook 10 Connecting knowledge volume 2

A box of seven blue cards numbered 1 through 7; five red cards numbered 1 to 5 and two yellow cards numbered 1 to 2 . Randomly draw a card.

a) Describe the sample space.

b) Which of the following is a subset of the sample space?

A: “Red or yellow card drawn”;

B: “Draw a card with a number of either 2 or 3”.

Solution method

– The symbol X is blue, D is red, V is yellow

– Find sample space

– Determine the elements of event A, event B

Detailed explanation

a) Sample space: \(\Omega\) = {X1; X2; X3; X4; X5; X6; X7; D1; D2; D3; D4; D5; V1; V2}

=> \(n(\Omega )\) = 14.

b)

A= {X1; X2; X3; X4; X5; X6; X7; D1; D2; D3; D4; D5}.

B = {X2; X3; D2; D3; V2}.

Solve problem 9.18 page 88 Math textbook 10 Connecting knowledge volume 2

There are boxes I and II, each containing 5 cards numbered 1 to 5. From each box, one card is drawn at random. Find the probability that the card drawn from box II has a greater number than the number on the card drawn from box I.

Solution method

– Find \(n(\Omega )\)

– Call Event A: “the card drawn from box II has a larger number than the number on the card drawn from box I”

– Calculate n(A) => P(A)

Detailed explanation

Draw from box I has 5 ways, from union II has 5 ways, the number of possibilities when drawing 1 card from each box is: 5.5 = 25, or \(n(\Omega )\) = 25.

	first	2	3	4	5
first	11	twelfth	13	14	15
2	21	22	23	24	25
3	thirty first	32	33	34	35
4	41	42	43	44	45
5	51	52	53	54	55

Event A: “The card drawn from box II has a larger number than the number on the card drawn from box I”.

A= {11; twelfth; 13 14; 15; 16; 23; 24; 25; 26; 34; 35; 36; 45; forty six; 56}.

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=> n(A) = 15

=> P(A) = \(\frac{15}{25}= \frac{3}{5}\).

Solve problem 9.19 page 88 Math textbook 10 Connecting knowledge volume 2

Roll two matching dice at the same time. Calculate the problem probability:

a) The sum of the dots on the two dice is 8;

b) The sum of the dots on the two dice is less than 8 .

Solution method

– Calculate \(n(\Omega )\)

– Call Event A: “The sum of the dots on the two dice is 8”. Calculate the number of favorable outcomes with A

– Call Event B: “The sum of the dots on the two dice is less than 8”.

+ Calculate the number of dots on the second dice when the numbers of the first dice are 1, 2, 3, 4, 5, 6 respectively.

Detailed explanation

Roll two dice, so the number of possible outcomes is: 6.6 = 36, or \(n(\Omega )\) = 36.

a) Event A: “The sum of the dots on the two dice is 8”.

There are 8 = 2 + 6 = 3 + 5 = 4 + 4. So the number of favorable outcomes for A is: 5.

P(A) = \(\frac{5}{36}\).

b) Event B: “The sum of the dots on the two dice is less than 8”.

+ If the number of dots on the first dice is 1, the number of dots on the second dice can be from 1 to 6: there are 6 ways.

+ If the number of dots on the first dice is 2, the number of dots on the second dice can be from 1 to 5: there are 5 ways.

+ If the number of dots on the first dice is 3, the number of dots on the second dice can be from 1 to 4: there are 4 ways.

+ If the number of dots on the first dice is 4, the number of dots on the second dice can be from 1 to 3: there are 3 ways.

+ If the number of dots on the first dice is 5, the number of dots on the second dice can be from 1 to 2: there are 2 ways.

+ If the number of dots on the first dice is 6, the number of dots on the second dice can be from 1: there is 1 way.

=> The number of ways is: 6+5+4+3+2+1 = 21 ways, or n(B) = 21.

=> P(B) = \(\frac{21}{36}=\frac{7}{12}\).

Solution 9.20 page 89 Math textbook 10 Connecting knowledge volume 2

The weather forecast for the next three days Monday, Tuesday, Wednesday of the following week says, in each of these days, the possibility of rain and no rain is the same.

a) Draw a tree diagram describing the sample space.

b) Calculate the probability of the events:

F: “In three days, there is exactly one rainy day”;

G: “For three days, there are at least two days without rain.”

Solution method

– Draw a tree diagram, symbol: M is rain, KM is no rain => \(n(\Omega )\)

– Based on the diagram to determine the number of elements of the events F, G. From there we find P(F), P(G)

Detailed explanation

a) The symbol M is rain, KM is no rain.

\(n(\Omega )\)= 8.

b)

+ Event F:

According to the diagram, n(F) = 3

=> P(F) = \(\frac{3}{8}\).

+ Event G:

According to the diagram, n(G) = 4

=> P(F) = \(\frac{4}{8}=\frac{1}{2}\).

Solution 9.21 page 89 Math textbook 10 Connecting knowledge volume 2

Toss a balanced coin four times in a row.

a) Draw a tree diagram describing the sample space.

b) Find the probability that in those four toss there are two tails and two heads.

Solution method

– Draw a tree diagram, symbol: S is heads, N is heads => \(n(\Omega )\)

– Based on the diagram to determine the number of elements of the events A. From there we find P(A)

Detailed explanation

a) The symbol S is for heads, N is for heads.

\(n(\Omega )\) = 16.

b) Event A: “In those four tosses there were two tails and two heads.”

n(A) = 6

=> P(A) = \(\frac{6}{16}=\frac{3}{8}\).

Solve problem 9.22, page 89 Math textbook 10 Connecting knowledge volume 2

Randomly select 4 marbles from a bag containing 4 different red and 6 blue marbles. Let A be the event: “In those four marbles, there are both red and blue marbles”. Calculate P(A) and P(\(\overline{A}\)).

Solution method

– Choose 4 marbles from 10 balls => \(n(\Omega )\)

– Calculate the number of ways to choose with the following cases:

+ Case 1: there is 1 green, 3 red

+ Case 2: there are 2 green, 2 red

+ Case 3: there are 3 green, 1 red

– Use the addition rule

Detailed explanation

Choose 4 marbles from 10 marbles, then the number of ways is: \(C_{10}^{4}\)= 210 ways.

=> \(n(\Omega )\) = 210.

Considering the event A, for both red and green, there are the following cases:

+ Case 1: there is 1 green, 3 red, the number of ways is: 6.\(C_{4}^{3}\) = 24

+ Case 2: there are 2 green, 2 red, the number of ways is: \(C_{6}^{2}.C_{4}^{2}\) = 90.

+ Case 3: there are 3 green, 1 red, the number of ways is: \(C_{6}^{3}\).4 = 80.

=> n(A) = 24+90+80 = 194.

=> P(A) = \(\frac{194}{210}= \frac{97}{105}\).

=> P(\(\overline{A}\)) = 1 – P(A) = \(\frac{8}{105}\).