## Solution of Exercise 5: Probability of the event (C6 – Math 10 Kite)——————

### Solution of Exercises Lesson 1, page 52, Math Textbook 10, Kite episode 2

A box has 5 cards of the same type, each card is marked with one of the numbers 1, 2, 3, 4, 5, two different cards have two different numbers. Randomly draw 2 cards from the box at the same time.

a) Let $$\Omega$$ be the pattern space in the above game. Calculate the number of elements of the set $$\Omega$$.

b) Calculate the probability of the event “The product of the numbers on two cards is odd”.

Solution method

a) Randomly draw 2 cards from the 5 cards in the box at the same time $$\Rightarrow$$Use combo formula

b) +) Step 1: Calculate the number of elements of the outcome in favor of the event “$$n\left( A \right)$$”

+) Step 2: The probability of the event is: $$P\left( A \right) = \frac{{n\left( A \right)}}{{n\left( \Omega \right)}}$$

Solution guide

a) Each element of the sample space is a convolution of 2 of 5 elements. Thus, the number of elements of the sample space is: $$n\left( \Omega \right) = C_5^2$$ (elements)

b)

+) Let A be the event “The product of the numbers on two cards is odd”

+) For the product of the numbers on the card to be odd, both cards drawn must be odd. Thus, the number of elements favorable outcomes for event A is a convolution 2 of 3 elements: $$n\left( A \right) = C_3^2$$ (elements)

+) So the probability of event A is: $$P\left( A \right) = \frac{{n\left( A \right)}}{{n\left( \Omega \right)}} = \frac{{C_3^2}}{{C_5^2}} = \frac{3}{{10}}$$

### Solution of Exercises Lesson 2, page 52, Math Textbook 10 Kite, episode 2

A box has 4 cards of the same type, each card is marked with one of the numbers 1, 2, 3, 4. Two different cards have two different numbers. Randomly draw 3 cards from the box at the same time.

a) Calculate the number of elements of the sample space.

b) Identify the following events:

A: “The sum of the numbers on the three cards is 9”;

B: “The numbers on the three cards are three consecutive natural numbers”.

c) Calculate P(A), P(B).

Solution method

a) Randomly draw 3 cards at the same time from 4 cards in the box $$\Rightarrow$$Use the combination formula

b) List cases in favor of events

c) The probability of the event is: $$P\left( A \right) = \frac{{n\left( A \right)}}{{n\left( \Omega \right)}};P\ left( B \right) = \frac{{n\left( B \right)}}{{n\left( \Omega \right)}}$$

Solution guide

a) Each element of the sample space is a convolution of 3 of 4 elements. Thus, the number of elements of the sample space is: $$n\left( \Omega \right) = C_4^3$$ (elements)

b) +) The event “The sum of the numbers on the three cards is 9” corresponds to the event $$A = \left\{ {\left( {4;3;2} \right)} \right\}\ ) +) The event “The numbers on the three cards are three consecutive natural numbers” corresponds to the event \(B = \left\{ {\left( {1;2;3} \right),\left( {2;3;4} \right)} \right\}$$

c) +) We have: $$n\left( A \right) = 1$$,$$n\left( B \right) = 2$$

+) Then the probability of events A and B is: $$P\left( A \right) = \frac{{n\left( A \right)}}{{n\left( \Omega \right)} } = \frac{1}{4};P\left( B \right) = \frac{{n\left( B \right)}}{n\left( \Omega \right)}} = \frac {2}{4} = \frac{1}{2}$$

### Solving exercises Lesson 3, page 52, Math Textbook 10 Kite, episode 2

Two female friends Hoa, Thao and two boys Dung and Huy were randomly assigned to sit in four chairs placed in a vertical line. Calculate the probability of each event:

a) “Friend Thao takes the first seat”;

b) “Friend Thao takes the first seat and friend Huy takes the last seat”.

Solution method

Step 1: Calculate the number of elements of the pattern space “$$n\left( \Omega \right)$$” and the number of elements of the result in favor of the event “$$n\left( A \right)\ )” Step 2: The probability of the event is: \(P\left( A \right) = \frac{{n\left( A \right)}}{{n\left( \Omega \right)}}$$

Solution guide

+) Putting 4 friends in 4 chairs is a permutation of 4 elements. Therefore, the pattern space is: $$n\left( \Omega \right) = 4!$$ (element)

a) +) Let A be the event “Friend Thao takes the first seat”

The first chair is Thao’s chair, so there is 1 way to choose, the remaining 3 seats are arranged arbitrarily 3 friends, so we have a permutation of 3 elements. According to the multiplication rule, we have: $$n\left( A \right) = 1.3!$$ ( element)

+) So the probability of event A is: $$P\left( A \right) = \frac{{n\left( A \right)}}{{n\left( \Omega \right)}} = \frac{1}{4}$$

b) +) Let B be the event “Friend Thao takes the first seat and friend Huy takes the last seat”.

Thao’s first chair and Huy’s last chair should have a way to choose for both seats, the remaining 2 seats are arranged arbitrarily by 2 friends, so we have a permutation of 2 elements. According to the multiplication rule, we have: $$n\left( B \right) = 1.1.2!$$ ( element)

+) Then the probability of event B is: $$P\left( B \right) = \frac{{n\left( B \right)}}{{n\left( \Omega \right)}} = \frac{1}{{12}}$$

### Solve exercises Lesson 4, page 52, Math textbook 10 Kite episode 2

There are 10 white flowers, 10 yellow flowers and 10 red flowers. People choose 4 flowers from the flowers above. Find the probability of the event “The four chosen flowers have all three colors”.

Solution method

+) Step 1: Calculate the number of elements of the sample space “$$n\left( \Omega \right)$$” and the number of elements of the result in favor of the event “$$n\left( A \right )$$” where A is the event “Four chosen flowers have all three colors”.

+) Step 2: The probability of the event is: $$P\left( A \right) = \frac{{n\left( A \right)}}{{n\left( \Omega \right)}}$$

Solution guide

+) Each time 4 flowers are randomly drawn from 30 flowers, we have a convolution of 4 of 30. Therefore, the number of elements of the sample space is: $$n\left( \Omega \right) = C_{ 30}^4$$ (element)

+) Let A be the event “four flowers are chosen with all three colors”

+) To choose four flowers with all 3 colors, we divide into three cases:

TH1: 2 white flowers, 1 yellow flower, 1 red flower: $$C_{10}^2.10.10$$ (how to choose)

TH2: 1 white flower, 2 yellow flowers, 1 red flower: $$10.C_{10}^2.10$$ (how to choose)

TH3: 1 white flower, 1 yellow flower, 2 red flowers: $$10.10.C_{10}^2$$ (how to choose)

+) Applying the addition rule, we have $$n\left( A \right) = 13500$$ (how to choose)

+) The probability of event A is: $$P\left( A \right) = \frac{{n\left( A \right)}}{{n\left( \Omega \right)}} = \ frac{{100}}{{203}}$$