Question

from a box containing two apples two peaches and two oranges four fruits are drawn at random. let P be a random variable representing the number of oranges that occur. construct a probability distribution

Answer 1

There are 6 fruits in total. We pick up 4 fruits. There are $C(6,4)$ ways to do it.

$$\displaystyle C(6,4)=\dbinom{6}{4}=\frac{6!}{4!(6-4)!} = \frac{ 5 \cdot 6}{1 \cdot 2} = 15$$

Let $P$ be a random variable representing the number of oranges that occur. To built probability distribution, we need to find following probabilities $P(P=0), P(P=1), P(P=2),$ because we have only 2 oranges.

$$P(P=0)=\dfrac{\dbinom{2}{0}\dbinom{6-2}{2-0}}{\dbinom{6}{4}}=\dfrac{1(6)}{15}=\dfrac{2}{5}$$

$$P(P=1)=\dfrac{\dbinom{2}{1}\dbinom{6-2}{2-1}}{\dbinom{6}{4}}=\dfrac{2(4)}{15}=\dfrac{8}{15}$$

$$P(P=2)=\dfrac{\dbinom{2}{2}\dbinom{6-2}{2-2}}{\dbinom{6}{4}}=\dfrac{1(1)}{15}=\dfrac{1}{15}$$

Construct a probability distribution

$$\def\arraystretch{1.5} \begin{array}{c:c} P & 0 & 1 & 2 \\ \hline p & 2/5 & 8/15 & 1/15 \\ \end{array}$$