Question

Suppose that a box contains seven red balls and three blue balls. If five balls are selected at random, without replacement, determine the p.f. of the number of red balls that will be obtained.

Short answer

The required probability function is,

\(f\left( x \right) = \frac{{\left( \begin{array}{l}7\\x\end{array} \right)\left( \begin{array}{l}\;\;3\\5 - x\end{array} \right)}}{{\left( \begin{array}{l}10\\5\end{array} \right)}}\;\;for\;x = 2,3,4,5,\)

Step-by-step solution

Given information

The number of red balls contained in a box is 7.

The number of blue balls contained in a box is 3.

The number of randomly selected balls without replacement is 5.

Determine the probability function

The total number of balls in a box is,

\(\begin{array}{c}n = 7 + 3\\ = 10\end{array}\).

Let X be the random variable representing the number of red balls in draw of 5 balls.

The total number of ways of selecting any random ball is\(\left( \begin{array}{l}10\\5\end{array} \right)\).

Out of 5 selected balls, at least 2 balls would be red. Thus, X can take 4 possible values, which are 2,3,4,5.

For x selected red ball, the number of favourable outcomes are,\(\left( \begin{array}{l}7\\x\end{array} \right)\left( \begin{array}{l}\;\;3\\5 - x\end{array} \right)\).

For the provided scenario, the probability function of X is,

\(f\left( x \right) = \frac{{\left( \begin{array}{l}7\\x\end{array} \right)\left( \begin{array}{l}\;\;3\\5 - x\end{array} \right)}}{{\left( \begin{array}{l}10\\5\end{array} \right)}}\;\;for\;x = 2,3,4,5,\)