An urn contains 5 red and 2 black balls. Two balls

Question 1:- An urn contains 5 red and 2 black balls. Two balls are randomly drawn, without replacement. Let X represent the number of black balls drawn. What are the possible values of X ? Is X is random variable ? If yes, find the mean of X.

Solution:- Let X = represent the number of black balls drawn

Number of red balls = 5

Number of black balls = 2

Possible value of X = 0, 1, 2

Yes X is a random variable

Now, $P(X = 0) = P(b)\times P(b)$

$= \dfrac{5}{7}\times \dfrac{4}{6} = \dfrac{20}{42}$

$P(X = 1) = P(b)\times P(r) + P(r)\times P(b)$

$=\dfrac{5}{7}\times \dfrac{2}{6} + \dfrac{2}{7}\times \dfrac{5}{6}$

$= \dfrac{10}{42} + \dfrac{10}{42} = \frac{20}{42}$

$P(X = 2) =P(r)\times P(r)$

$=\dfrac{2}{7}\times \dfrac{1}{6} = \dfrac{2}{42}$

Table for calculation of mean

$\begin{array}{|c|c|c|c|}\hline X & 0 & 1 & 2 \\\hline P(X) & \frac{20}{42} & \frac{20}{42} & \frac{2}{42} \\ \hline \end{array}$

Mean = ∑P(X) $= 0\times \dfrac{20}{42}+1\times \dfrac{20}{42}+2\times \dfrac{2}{42}$

$= \dfrac{20}{42}+\dfrac{4}{42}$

$= \dfrac{24}{42}=\dfrac{4}{7}$

Question 2:- An urn contains 4 white and 3 red balls. Let X be the number of red balls in random draw of three balls. Find the mean of X.

Solution:- Let X = The number of red balls in random draw of three balls.

As there are 3 red balls, possible value of X = 0, 1, 2, 3

$P(X =0) =P(W)\times P(W) \times P(W)$

$=\dfrac{4}{7}\times \dfrac{3}{6}\times \dfrac{2}{5}$

$= \dfrac{4}{35}$

$P(X=1) = P(R)\times P(W) \times P(W)+P(R)\times P(W) \times P(R)+P(R)\times P(R) \times P(W)$

$=\dfrac{3}{7}\times \dfrac{4}{6}\times \dfrac{3}{5} + \dfrac{4}{7}\times \dfrac{3}{6}\times \dfrac{3}{5}+\dfrac{4}{7}\times \dfrac{3}{6}\times \dfrac{3}{5}$

$=\dfrac{18}{35}$

$P(X = 2) = P(R)\times P(R) \times P(W)+P(R)\times P(W) \times P(R)+ P(W)\times P(R) \times P(R)$

$= \dfrac{3}{7} \times \dfrac{2}{6}\times \dfrac{4}{5}+\dfrac{3}{7} \times \dfrac{4}{6}\times \dfrac{2}{5}+\dfrac{4}{7} \times \dfrac{3}{6}\times \dfrac{2}{5}$

$= \dfrac{12}{35}$

$P(X = 3) = P(R)\times P(R) \times P(R)$

$= \dfrac{3}{7}\times \dfrac{2}{6}\times \dfrac{1}{5}$

$= \dfrac{1}{35}$

Probability distribution

$\begin{array}{|c|c|c|c|c|}\hline X & 0 & 1 & 2 & 3 \\\hline P(X) & \frac{4}{35} & \frac{18}{35} & \frac{12}{35} & \frac{1}{35}\\ \hline \end{array}$

Mean = ∑P(X) $= 0\times \dfrac{4}{35}+1\times \dfrac{18}{35}+2\times \dfrac{12}{35}+ 3\times \dfrac{1}{35}$

$= 0 + \dfrac{18}{35} + \dfrac{24}{35}+\dfrac{3}{35}$

$= \dfrac{45}{35}=\dfrac{9}{7}$

Question:- In a certain collage, 4% of boys and 1% of girls are taller than 1.75 metres, Furthermore, 60% of the students in the college are girls. A student is selected at random from the college and is found to be taller than 1.75 metres. Find the probability that the selected student is a girl.

Solution:- See full solution