adsense
Solving Exercises Lesson 26: Events and the classical definition of probability (Math 10 – Connecting Textbook)
——————
Solve lesson 9.1, page 82, Math 10 Textbook, Connecting knowledge volume 2
Randomly select a positive integer not greater than 30.
a) Describe the sample space.
b) Let A be the event: “The number chosen is prime”. The events A and \(\overline{A}\) are which subset of the sample space?
Solution method
a) Define the pattern space \(\Omega \)
b) determine the event A, calculate \(\overline{A}\)
Detailed explanation
a) Sample space \(\Omega \) = {1; 2; 3; 4; 5; 6; 7; 8; 9; ten; 11; twelfth; 13; 14; 15; 16; 17; 18; 19; 20; 21; 22; 23 ;24; 25; 26 ; 27; 28; 29; 30}.
b) A = {2; 3; 5; 7; 11; 13; 17; 19; 23; 29}
\(\overline{A}\) = {1; 4; 6; 8; 9; ten; twelfth; 14; 15; 16; 18; 20; 21; 22; 24; 25; 26; 27; 28; 30}.
Solve problem 9.2, page 82, Math textbook 10, Connecting knowledge volume 2
Randomly choose a positive integer not greater than 22.
a) Describe the sample space.
b) Let B be the event: “The selected number is divisible by 3”. The events B and $\overline{B}$ are what subsets of the sample space?
Solution method
a) Define the pattern space \(\Omega \)
b) determine event B, calculate \(\overline{B}\)
Detailed explanation
a) Sample space \(\Omega \) = {1; 2; 3; 4; 5; 6; 7; 8; 9; ten; 11; twelfth; 13; 14; 15; 16; 17; 18; 19; 20; 21; 22}.
b) B = {3; 6; 9; twelfth; 15; 18; 21}
\(\overline{B}\) = {1; 2; 4; 5; 7; 8; ten; 11; 13; 14; 16; 17; 19; 20; 22}.
Solve lesson 9.3 page 82 Math textbook 10 Connecting knowledge volume 2
Roll a dice and a coin at the same time.
a) Describe the sample space.
b) Consider the following events:
C: “The coin appears tails”;
D: “The coin that appears heads or the number of dots on the dice is 5”.
The events $C, \overline{C}, D$, and $\overline{D}$ are what subsets of the sample space?
Solution method
– Tabulate the sample space => \(n(\Omega )\)
– Determine C, D. Calculate \(\overline{C}\), \(\overline{D}\)
Detailed explanation
a) The symbol S is for heads, N is for heads. The sample space is given according to the table:
| first | 2 | 3 | 4 | 5 | 6 | |
| S | S1 | S2 | S3 | S4 | S5 | S6 |
| WOMEN | N1 | N2 | N3 | N4 | N5 | N6 |
So \(n(\Omega )\) = 10.
b)
C = {S1; S2; S3; S4; S5; S6}
adsense
\(\overline{C}\) = {N1; N2; N3; N4; N5; N6}
D = {N1; N2; N3; N4; N5; N6; S5}
\(\overline{D}\) = {S1; S2; S3; S4; S6}
Solve problem 9.4 page 82 Math textbook 10 Connecting knowledge volume 2
A bag contains a number of blue marbles, red marbles, black marbles and white marbles. Randomly pick a marble from the bag.
a) Let H be the event: “The ball drawn is red”. Event: “Is the ball drawn blue or black or white” an event \(\overline{H}\) or not?
b) Let K be the event: “The marbles drawn are blue or white”. Event: “Bit drawn is black” is the event \(\overline{K}\) or not?
Solution method
The opposite event of event E is the event “E does not occur”. The opposite event of E is denoted by \(\overline E \).
Detailed explanation
a) Event: “The marble drawn is blue or black or white” is the event \(\overline{H}\) because if the red marble is not drawn, it can only be blue or black, or white. .
b) Event: “Black drawn marble” is not an event \(\overline{K}\) because if blue or white is not drawn, it could be black or red.
Solve problem 9.5 page 82 Math textbook 10 Connecting knowledge volume 2
Two friends An and Binh each roll a balanced dice. Calculate the probability that:
a) The number of dots appearing on the two dice is less than 3;
b) The number of dots appearing on the dice that An rolls is greater than or equal to 5;
c) The product of two numbers appearing on two dice less than 6;
d) The sum of the two numbers appearing on the two dice is a prime number.
Solution method
– Calculate sample space
– Determine event elements A=> \(P(A)=\frac{n(A)}{n(\Omega )}\)
Similarly Determine event elements B, C, D => \(P(B)\), \(P(C)\), \(P(D)\)
Detailed explanation
Because rolling a dice, the number of dots that can appear is 1, 2, 3, 4, 5, 6, so when rolling 2 dice, the number of possibilities is \(n(\Omega )\) = 6.6 = 36.
The results of the sample space are given in the table:
| first | 2 | 3 | 4 | 5 | 6 | |
| first | (1;1) | (1;2) | (1;3) | (1;4) | (1;5) | (1;6) |
| 2 | (2;1) | (2;2) | (2;3) | (2;4) | (2;5) | (2;6) |
| 3 | (3;1) | (3;2) | (3;3) | (3;4) | (3;5) | (3;6) |
| 4 | (4;1) | (4;2) | (4;3) | (4;4) | (4;5) | (4;6) |
| 5 | (5;1) | (5;2) | (5;3) | (5;4) | (5;5) | (5,6 |
| 6 | (6;1) | (6;2) | (6;3) | (6;4) | (6;5) | (6;6) |
a) Event A: “The number of dots appearing on the two dice is less than 3”.
The favorable outcomes of A are: (1;1), (1,2), (2,1), (2,2).
n(A) = 4. So \(P(A)=\frac{n(A)}{n(\Omega )}=\frac{4}{36}=\frac{1}{9}\) .
b) Event B: “The number of dots appearing on the dice An rolled is greater than or equal to 5”.
The favorable outcomes of B are:
(5;1), (5,2), (5;3), (5;4), (5,5), (5,6), (6,1), (6,2), (6 ;3), (6,4), (6,5), (6,6).
n(B) = 12. So \(P(B)=\frac{n(B)}{n(\Omega )}=\frac{12}{36}=\frac{1}{3}\) .
c) Event C: “The product of two numbers appearing on two dice is less than 6”.
The favorable outcomes of C are: (1; 1), (1; 2), (1; 3), (1; 4), (1; 5), (2; 1), (3; 1) , (4; 1), (5; 1).
n(C) = 9. So \(P(C)=\frac{n(C)}{n(\Omega )}=\frac{9}{36}=\frac{1}{4}\) .
d) Event D: “The sum of the two numbers appearing on the two dice is a prime number”.
The favorable outcomes of D are: (1; 1), (1; 2), (2; 1), (1; 4), (4; 1), (1; 6), (6, 1) , (2; 3); (2; 5), (3; 2), (5; 2), (3; 4), (4; 3), (5; 6), (6; 5).
n(D) = 15. So \(P(D)=\frac{n(D)}{n(\Omega )}=\frac{15}{36}=\frac{5}{12}\) .