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**Solution of Exercise 27: Practice calculating probability according to classical definition (Math 10 – Connecting textbook)**

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### Solve lesson 9.6, page 86, Math textbook 10, Connecting knowledge, volume 2

Randomly select a family with three children and observe the sexes of these three children. Calculate the probability of the following events:

a) A: “The first child is a girl”;

b) B: “Has at least one son”.

**Solution method**

– Find \(n(\Omega )\)

– Find the number of ways to choose the first child, the number of ways to choose the next two children-

– consider and find the event \(\overline{B}\): “There are no sons”.

**Detailed explanation**

Each child will be either a boy or a girl, so 3 children, the number of possibilities is: 2.2.2 = 8, or \(n(\Omega )\) = 8.

a) The first child is a girl, so there is only one choice.

The following two children, regardless of gender, should have: 2.2 = 4 choices.

=> n(A) = 1.4 = 4. So P(A) = \(\frac{4}{8}=\frac{1}{2}\).

b) consider the event \(\overline{B}\): “There are no sons”.

In order for there to be no boys, then all three children are girls, so n(\(\overline{B}\)) = 1.

=> P(\(\overline{B}\)) = \(\frac{1}{8}\)

=> P(B) = 1- P(\(\overline{B}\)) = \(\frac{7}{8}\)

### Solve problem 9.7 page 86 Math textbook 10 Connecting knowledge volume 2

A box of cards numbered 10; 11; ….; 20. Randomly draw two cards from a box. Calculate the probability of the following events:

a) C: “Both draw cards have odd numbers”;

b) D: “Both cards drawn have even numbers”.

**Solution method**

Calculate \(n(\Omega )\)

– The two cards drawn are odd numbers, so the two cards drawn belong to set {11; 13; 15; 17; 19} => Number of ways to choose

– The two cards drawn have even numbers, so the two cards drawn belong to set {10; twelfth; 14; 16; 18; 20} => Number of ways to choose

**Detailed explanation**

Draw two cards from 11 cards with the number of ways: \(C_{11}^{2}=55\) or \(n(\Omega )\) = 55.

a) Both cards drawn are odd numbers, so 2 cards drawn belong to set {11; 13; 15; 17; 19}.

=> The number of ways to choose is: \(C_{5}^{2}=10\).

So P(C) = \(\frac{10}{55}=\frac{2}{11}\).

b) Both cards drawn have even numbers, so 2 cards drawn belong to set {10; twelfth; 14; 16; 18; 20}

=> Number of ways to choose is: \(C_{6}^{2}=15\).

So P(D) = \(\frac{15}{55}=\frac{3}{11}\).

### Solve problem 9.8, page 86, Math textbook 10, Connecting knowledge volume 2

A box contains 6 white marbles, 4 red marbles and 2 black marbles. Randomly select 6 marbles. Find the probability that there are 3 white marbles, 2 red marbles, and 1 black marble among those 6.

**Solution method**

– Find \(n(\Omega )\) choose 6 marbles out of 12

Number of ways to choose 3 white marbles out of 6

Number of ways to choose 2 red marbles out of 4

Number of ways to choose 1 black marble out of 2

**Detailed explanation**

If you choose 6 marbles out of 12, the number of ways to choose is: \(C_{12}^{6}\) = 924 ways, or \(n(\Omega )\) = 924.

Event A: “In those 6 marbles there are 3 white marbles, 2 red marbles and 1 black marble”.

+ Choose 3 white marbles out of 6, number of ways: \(C_{6}^{3}\) = 20.

+ Choose 2 red marbles out of 4, number of ways: \(C_{4}^{2}\) = 6.

+ Choose 1 black marble out of 2, number of ways: \(C_{2}^{1}\) = 2.

=> n(A) = 20.6.2 = 240

So P(A) = \(\frac{240}{924}=\frac{20}{77}\).

### Solve problem 9.9 page 86 Math textbook 10 Connecting knowledge volume 2

Roll a dice and a coin in a row.

a) Draw a tree diagram describing the elements of the sample space.

b) Calculate the probability of the following events:

F: “The coin appears heads”;

G: “The coin that appears heads or the number of dots on the dice is 5”.

**Solution method**

+ Symbol S is heads, N is heads => \(n(\Omega )\)

+ Find event F, favorable outcomes for event F

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+ Find event G, favorable outcomes for event G

**Detailed explanation**

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

\(n(\Omega )\) = 12

b)

+ Event F, favorable outcomes for event F are: {N1; N2; N3; N4; N5; N6}.

=> n(F) = 6

=> P(F) = \(\frac{6}{12}=\frac{1}{2}\).

+ Event G, the favorable outcomes for event G are: {S1; S2; S3; S4; S5; S6; N5}.

=> n(G) = 7

=> P(G) = \(\frac{7}{12}\).

### Solve lesson 9.10, page 87, Math textbook 10, Connecting knowledge volume 2

On a street, there are two restaurants X, Y. Three friends Son, Hai and Van each choose a restaurant at random.

a) Draw a tree diagram describing the elements of the sample space.

b) Calculate the probability of the event “Two friends go to shop X, the other friend goes to shop Y”.

**Solution method**

– Draw a tree diagram to find \(n(\Omega )\)

– Find a favorable outcome for event A

**Detailed explanation**

a)

\(n(\Omega )\)= 6.

b) Event A: “Two friends go to shop X, the other friend goes to shop Y”.

Favorable outcomes for event A: {XXY; XYX; YXX}

=> n(A) = 3

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

### Solve problem 9.11 page 87 Math textbook 10 Connecting knowledge volume 2

Roll two balanced dice in turn. Find the probability that at least one of the dice will have a 6-dot face.

**Solution method**

– Find the sample space: \(n(\Omega )\)

– Call event A: “at least one dice appears 6 dots”

+ Case: a 6-dot, a non-6-dot, number of possibilities.

+ Case: both children 6 dots, number of possibilities.

**Detailed explanation**

Sample space: \(n(\Omega )\) = 6.6 = 36.

Consider event A: “at least one dice appears with a six-dot face”

For at least one dice to appear on a 6-dot face, the possibilities are:

+ Case: one is 6 dots, one is not 6 dots, number of possibilities: 1.6. 2 = 12

+ Case: both children 6 dots, number of possibilities: 1.

=> n(A) = 13

=> P(A) = \(\frac{13}{36}\)

### Solve problem 9.12 page 87 Math textbook 10 Connecting knowledge volume 2

The seed color of peas has two phenotypes, yellow and green, corresponding to two genes, dominant gene A and recessive gene a. The seed shape of peas has two phenotypes, smooth and wrinkled, corresponding to two types of genes: dominant gene B and recessive gene b. Assume that the seedlings randomly pick one gene from the parent plant and one gene from the parent plant.

The test is to cross two types of peas, in which both the parent plant and the parent plant have the genotype (Aa,Bb) and the phenotype is yellow and smooth seeds. Assume the possible outcomes are congruent. Find the probability that the seedling also has a yellow and smooth seed phenotype.

**Solution method**

– Find \(n(\Omega )\)

– Call event A: “the seedling also has a yellow and smooth seed phenotype.”

– Find favorable outcomes for event A

**Detailed explanation**

Samplespace \(\Omega \) = {AABB, AABb, AAbb, aabb, aaBB, aaBb, AaBB, AaBb, Aabb}

=> \(n(\Omega )\) = 9.

Event A: “The seedlings also have a yellow and smooth seed phenotype.”

In order for a seedling to have a smooth and yellow seed phenotype, A and B must be present in it. The outcomes are favorable for event A: {AABB, AABb, AaBB, AaBb}.

=> n(A) = 4

=> P(A) = \(\frac{4}{9}\).