Question

A foreign student club lists as its members 2 Canadians, 3 Japanese, 5 Italians, and 2 Germans. If a committee of 4 is selected at random, find the probability that (a) all nationalities are represented; (b) all nationalities except Italian are represented

Answer 1

(a) The probability that the members of the committee are chosen from all nationalities
`=(4)/(33)` =0.1212.

(b)The probability that all nationalities except Italian are represent is 0.04848.

Explanation:

Hypergeometric Distribution:

Let
`x_1`,
`x_2`,
`x_3` and
`x_4` be four given positive integers and let
`x_1+x_2+x_3+x_4= N`.

A random variable X is said to have hypergeometric distribution with parameter
`x_1`,
`x_2`,
`x_3` ,
`x_4` and n.

The probability mass function

`f(x_1,x_2.x_3,x_4;a_1,a_2,a_3,a_4;N,n)=\frac{\left(\begin{array}{c}x_1\\a_1\end{array}\right)\left(\begin{array}{c}x_2\\a_2\end{array}\right) \left(\begin{array}{c}x_3\\a_3\end{array}\right) \left(\begin{array}{c}x_4\\a_4\end{array}\right) }{\left(\begin{array}{c}N\\n\end{array}\right) }`

Here
`a_1+a_2+a_3+a_4=n`

`{\left(\begin{array}{c}x_1\\a_1\end{array}\right)=^(x_1)C_(a_1)= (x_1!)/(a_1!(x_1-a_1)!)`

Given that, a foreign club is made of 2 Canadian members, 3 Japanese members, 5 Italian members and 2 Germans members.

`x_1`=2,
`x_2`=3,
`x_3` =5 and
`x_4`=2.

A committee is made of 4 member.

N=4

(a)

We need to find out the probability that the members of the committee are chosen from all nationalities.

`a_1`=1,
`a_2`=1,
`a_3`=1 ,
`a_4`=1, n=4

The required probability is

`=\frac{\left(\begin{array}{c}2\\1\end{array}\right)\left(\begin{array}{c}3\\1\end{array}\right) \left(\begin{array}{c}5\\1\end{array}\right) \left(\begin{array}{c}2\\1\end{array}\right) }{\left(\begin{array}{c}12\\4\end{array}\right) }`

`=(2* 3* 5* 2)/(495)`

`=(4)/(33)`

=0.1212

(b)

Now we find out the probability that all nationalities except Italian.

So, we need to find out,

`P(a_1=2,a_2=1,a_3=0,a_4=1)+P(a_1=1,a_2=2,a_3=0,a_4=1)+P(a_1=1,a_2=1,a_3=0,a_4=2)`

`=\frac{\left(\begin{array}{c}2\\2\end{array}\right)\left(\begin{array}{c}3\\1\end{array}\right) \left(\begin{array}{c}5\\0\end{array}\right) \left(\begin{array}{c}2\\1\end{array}\right) }{\left(\begin{array}{c}12\\4\end{array}\right) }+\frac{\left(\begin{array}{c}2\\1\end{array}\right)\left(\begin{array}{c}3\\2\end{array}\right) \left(\begin{array}{c}5\\0\end{array}\right) \left(\begin{array}{c}2\\1\end{array}\right) }{\left(\begin{array}{c}12\\4\end{array}\right) }`
`+\frac{\left(\begin{array}{c}2\\1\end{array}\right)\left(\begin{array}{c}3\\1\end{array}\right) \left(\begin{array}{c}5\\0\end{array}\right) \left(\begin{array}{c}2\\2\end{array}\right) }{\left(\begin{array}{c}12\\4\end{array}\right) }`

`=(1* 3* 1* 2)/(495)+(2* 3* 1* 2)/(495)+(2* 3* 1* 1)/(495)`

`=(6+12+6)/(495)`

`=(8)/(165)`

=0.04848

The probability that all nationalities except Italian are represent is 0.04848.