Mathematics, 02.04.2020 02:04 FortniteB

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

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Answer from: earthangel456

A) The probability that all nationalities are represented is 0.5%.

B) The probability that all nationalities except Italian are represented is 7%.

Given that 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, to ﬁ nd the probability that (A) all nationalities are represented; (B) all nationalities except Italian are represented; the following calculations must be performed:

2 + 3 + 5 + 2 = 12

2/12 x 3/11 x 5/10 x 2/9 = X 0.0050 = X 0.0050 x 100 = 0.5

Therefore, the probability that all nationalities are represented is 0.5%.

7/12 x 6/11 x 5/10 x 4/9 = X 0.070 = X 0.070 x 100 = 7

In turn, the probability that all nationalities except Italian are represented is 7%.

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Answer from: stankyweezle

(a) The probability that the members of the committee are chosen from all nationalities $=\frac{4}{33}$ =0.1212.

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

Step-by-step 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

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

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

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

$=\frac{2\times 3\times 5\times 2}{495}$

$=\frac{4}{33}$

=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)

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

$=\frac{1\times 3\times 1\times 2}{495}+\frac{2\times 3\times 1\times 2}{495}+\frac{2\times 3\times 1\times 1}{495}$

$=\frac{6+12+6}{495}$

$=\frac{8}{165}$

=0.04848

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

Answer from: Quest

answer: x=60

Can someone plz answer this for me? whoever answers it first, i will give !

Answer from: Quest

It’s either a or c as the answer.

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