adsense
Solution of Exercise 5: Probability of the event (C6 – Math 10 Kite)
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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: \(n\left( A \right) = 1\),\(n\left( B \right) = 2\)
adsense
+) 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: \(P\left( A \right) = \frac{{n\left( A \right)}}{{n\left( \Omega \right)}} = \ frac{{100}}{{203}}\)
