## Solving Exercises Lesson 4: Probability of events in some simple games (C6 – Math 10 Kite)——————

### Solve the exercise Exercise 1, page 45, Math textbook 10 Kite, episode 2

Toss a coin twice in a row. Calculate the probability of the event “The outcome of the two tosses is different”.

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 “$$n\left( A \right )$$”

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

Solution guide

+) The pattern space in the above game is the set $$\Omega = {\rm{ }}\left\{ {SS;{\rm{ }}SN;{\rm{ }}NS;{\rm{ }}NN} \right\}$$. So $$n\left( \Omega \right) = 4$$

+) Let A be the event “The outcome of two tosses is different”.

The favorable outcomes for event A are: $$SN;{\rm{ }}NS$$i.e. $$A = \left\{ {SN;NS} \right\}$$.So $$n \left( A \right) = 2$$

+) So the probability of event A is: $$P\left( A \right) = \frac{{n\left( A \right)}}{{n\left( \Omega \right)}} = \frac{2}{4} = \frac{1}{2}$$

### Solution of Exercises Lesson 2, page 45, Math textbook 10 Kite, episode 2

Toss a coin three times in a row.

a) Write the set $$\Omega$$ as the sample space in the above game.

b) Identify each event:

Solution method

a) The sample space is all the possibilities of tossing the coin three times in a row

b) Based on the sample space to list

Solution guide

a) +) The sample space in the above game is the set

$$\Omega = {\rm{ }}\left\{ {SSS;{\rm{ }}SSN;{\rm{ }}SNN;{\rm{ }}SNS;{\rm{ }}NSN; {\rm{ }}NSS;NNS;{\rm{ }}NNN} \right\}$$

b) +) Event A is the set: $$A = \left\{ {NSN;{\rm{ }}NSS;NNS;{\rm{ }}NNN} \right\}$$ .

+) Event B is the set: $$B = \left\{ {SNS;{\rm{ }}SSN;NSS} \right\}$$

### Solve the exercises Exercise 3, page 45, Math textbook 10 Kite, episode 2

Roll a dice twice in a row. State each of the following events as fact clauses:

$$\begin{array}{*{20}{l}}{A = \left\{ {\left( {6;1} \right);\left( {6;2} \right);\left ( {6;3} \right);\left( {6;4} \right);\left( {6;5} \right);\left( {6;6} \right)} \right\} ;}\\{B = \left\{ {\left( {1;6} \right);\left( {2;5} \right);\left( {3;4} \right);\left ( {4;3} \right);\left( {5;2} \right);\left( {6;1} \right)} \right\};}\\{C = \left\{ { \left( {1;1} \right);\left( {2;2} \right);\left( {3;3} \right);\left( {4;4} \right);\left ( {5;5} \right);\left( {6;6} \right)} \right\}.}\end{array}$$

Solution method

Observe the common properties of each set

Solution guide

a) A is the event “To roll a dice twice in a row such that the first dice always appear green”

b) B is the event “Toss a dice twice in a row so that the total number of dots appearing is 7”

c) C is the event “Toss a dice twice in a row so that the number of dots appearing on two rolls is the same”

### Solving exercises Lesson 4, page 45, Math textbook 10 Kite, episode 2

Roll a dice twice in a row. Calculate the probability of each of the following events:

a) “The total number of dots appearing in two sows is not less than 10”;

b) “The 1-dot face appears at least once”.

Solution method

Step 1: Calculate the number of elements of the pattern space “$$n\left( \Omega )$$” and the number of elements of the result in favor of the event “$$n\left( A )$$,$$n\left( B )$$”

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

Solution guide

+) The pattern space in the above game is the set $$\Omega = {\rm{ }}\left\{ {\left( {i,j} ){\rm{ | }}i,{\rm{ }}j{\rm{ }} = {\rm{ }}1,{\rm{ }}2,{\rm{ }}3,{\rm{ }}4,{\rm{ }}5, {\rm{ }}6} \right\}$$ where (i,j) is the result “The first time appears the dotted face, the second time appears the dotted face j”. So $$n\left( \Omega ) = 36$$

a) Let A be the event “The total number of dots appearing in two sows is not less than 10”.

The outcomes in favor of A are: (4; 6) (5,5) (5,6) (6; 4) (6,5) (6,6). So $$n\left( A ) = 6$$

So the probability of event A is: $$P\left( A ) = \frac{{n\left( A )}}{{n\left( \Omega )}} = \frac{6}{{36 }} = \frac{1}{6}$$

b) Let B be the event “The dotted side appears at least once”

The results in favor of B are: (1; 1) (1 : 2) (1 : 3) (1; 4) (1;5) (1; 6) (2 ; 1) (3;1) ( 4; 1) (5;1) (6;1). So $$n\left( B ) = 11$$

So the probability of event A is: $$P\left( B ) = \frac{{n\left( B )}}{{n\left( \Omega )}} = \frac{{11}}{ {36}} = \frac{{11}}{{36}}$$