Answer to Question #168841 in Statistics and Probability for Ziah

Question

Construct the probability distribution and find the mean, variance and standard deviation of the following situation.

1. A group of friends is composed of 5 boys and 4 girls. You like to be off friend to 4 of them. The random variable T gives the number of girlfriends you can have.

Answer 1

The probability of having 0 girlfriends:

$p(0) = \frac{{C_5^4}}{{C_9^4}} = \frac{{5!}}{{4!1!}} \cdot \frac{{4!5!}}{{9!}} = 5 \cdot \frac{{2 \cdot 3 \cdot 4}}{{6 \cdot 7 \cdot 8 \cdot 9}} = \frac{5}{{7 \cdot 2 \cdot 9}} = \frac{5}{{126}}$

The probability of having 1 girlfriend:

$p(1) = \frac{{C_5^3C_4^1}}{{C_9^4}} = \frac{{5!}}{{3!2!}} \cdot \frac{{4!}}{{3!1!}} \cdot \frac{{4!5!}}{{9!}} = \frac{{4 \cdot 5}}{2} \cdot 4 \cdot \frac{{2 \cdot 3 \cdot 4}}{{6 \cdot 7 \cdot 8 \cdot 9}} = \frac{{40}}{{7 \cdot 2 \cdot 9}} = \frac{{20}}{{63}}$

The probability of having 2 girlfriends:

$p(2) = \frac{{C_5^2C_4^2}}{{C_9^4}} = \frac{{5!}}{{3!2!}} \cdot \frac{{4!}}{{2!2!}} \cdot \frac{{4!5!}}{{9!}} = \frac{{4 \cdot 5}}{2} \cdot \frac{{3 \cdot 4}}{2} \cdot \frac{{2 \cdot 3 \cdot 4}}{{6 \cdot 7 \cdot 8 \cdot 9}} = \frac{{60}}{{7 \cdot 2 \cdot 9}} = \frac{{30}}{{63}}$

The probability of having 3 girlfriends:

$p(3) = \frac{{C_5^1C_4^3}}{{C_9^4}} = \frac{{5!}}{{1!4!}} \cdot \frac{{4!}}{{3!1!}} \cdot \frac{{4!5!}}{{9!}} = 5 \cdot 4 \cdot \frac{{2 \cdot 3 \cdot 4}}{{6 \cdot 7 \cdot 8 \cdot 9}} = \frac{{20}}{{7 \cdot 2 \cdot 9}} = \frac{{10}}{{63}}$

The probability of having 4 girlfriends:

$p(4) = \frac{{C_4^4}}{{C_9^4}} = \frac{{4!5!}}{{9!}} = \frac{{2 \cdot 3 \cdot 4}}{{6 \cdot 7 \cdot 8 \cdot 9}} = \frac{1}{{7 \cdot 2 \cdot 9}} = \frac{1}{{126}}$

We have the probability distribution:

$\begin{matrix} T&0&1&2&3&4\\ p&{\frac{5}{{126}}}&{\frac{{20}}{{63}}}&{\frac{{30}}{{63}}}&{\frac{{10}}{{63}}}&{\frac{1}{{126}}} \end{matrix}$

Find the mean:

$M(T) = \sum {{t_i}} {p_i} = 0 \cdot \frac{5}{{126}} + 1 \cdot \frac{{20}}{{63}} + 2 \cdot \frac{{30}}{{63}} + 3 \cdot \frac{{10}}{{63}} + 4 \cdot \frac{1}{{126}} = \frac{{112}}{{63}} = \frac{{16}}{9}$

Find the variance:

$D(T) = M({T^2}) - {M^2}(T) = 0 \cdot \frac{5}{{126}} + 1 \cdot \frac{{20}}{{63}} + 4 \cdot \frac{{30}}{{63}} + 9 \cdot \frac{{10}}{{63}} + 16 \cdot \frac{1}{{126}} - {\left( {\frac{{16}}{9}} \right)^2} =$

$= \frac{{238}}{{63}} - \frac{{256}}{{81}} = \frac{{50}}{{81}}$

Find the standard deviation:

$\sigma (T) = \sqrt {D(T)} = \sqrt {\frac{{50}}{{81}}} = \frac{{5\sqrt 2 }}{9}$