## Solution 44 Page 50 Math Workbook 10 – Kite>

Topic

An international workshop includes 12 students from the following countries: Vietnam, Japan, Singapore, India, Korea, Brazil, Canada, Spain, Germany, France, South Africa, Cameroon, each country has exactly 1 the student. Randomly select 2 students from the international student group to participate in BTC:

Calculate the probability of each of the following events:

a) A: “Two selected students are from Asia”

b) B: “Two selected students from Europe”

c) C: “Two selected students from America”

d) D: “Two selected students from Africa”

Solution method – See details

The probability of event A being a number, symbol $$P\left( A \right)$$ is determined by the formula: $$P\left( A \right) = \frac{{n\left( A \right)}}{{n\left( \Omega \right)}}$$, where $$n\left( A \right)$$ and $$n\left( \Omega \right)$$ denote the number of elements of set A and $$\Omega$$ respectively.

Detailed explanation

Select 2 students from 12 students $$\Rightarrow$$ convolution 2 of 12 $$\Rightarrow n\left( \Omega \right) = C_{12}^2 = 66$$

a) A: “Two selected students from Asia”:

There are 5 Asian countries: Vietnam, Japan, Singapore, India, HQ

$$\Rightarrow n\left( A \right) = C_5^2 = 10$$

$$\Rightarrow P\left( A \right) = \frac{{n\left( A \right)}}{{n\left( \Omega \right)}} = \frac{{10}}{{ 66}} = \frac{5}{{33}}$$

b) B: “Two selected students from Europe”: There are 3 European countries: Spain, Germany, France $$\Rightarrow n\left( B \right) = C_3^2 = 3$$

$$\Rightarrow P\left( B \right) = \frac{{n\left( B \right)}}{{n\left( \Omega \right)}} = \frac{3}{{66} } = \frac{1}{{22}}$$

c) C: “Two selected students from America”: There are 2 American countries: Brazil, Canada $$\Rightarrow n\left( C \right) = C_2^2 = 1$$

$$\Rightarrow P\left( C \right) = \frac{{n\left( C \right)}}{{n\left( \Omega \right)}} = \frac{1}{{66} }$$

d) D: “Two selected students from Africa”: There are 2 African countries: South Africa, Cameroon $$\Rightarrow n\left( D \right) = C_2^2 = 1$$

$$\Rightarrow P\left( D \right) = \frac{{n\left( D \right)}}{{n\left( \Omega \right)}} = \frac{1}{{66} }$$