Question

From a box containing 2 apples, 2 peaches and 2 oranges, 3 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 3 fruits. There are $C(6,3)$ ways to do it.

$\displaystyle C(6,3)=\frac{6!}{3!\,3!} = \frac{4 \cdot 5 \cdot 6}{3 \cdot 2} = 20$

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). P=3 and more is not possible, because we have only 2 oranges.

P=0. We can choose any fruits, but not oranges, therefore, the number of ways to do it is $C(4,2) =6$.

$\displaystyle P(P=0) = \frac{6}{20} = \frac{3}{10} = 0.3$

P=1. We need to pick up exactly one orange. So the amount of possible combinations where 1 fruit is orange is equal to the number of ways we can choose 2 fruits out of 5.

$\displaystyle P(P=1) = \frac{C(5,2)}{20} = \frac{10}{20} = 0.5$

P=2. We have chosen 2 oranges and need to add only one fruit from the rest 4 fruits.$\displaystyle P(P=2) =\frac{ C(4,1)}{20} = \frac{4}{20} = 0.2$

Check: P(P=0) + P(P=1) + P(P=2) = 0.3 + 0.5 +0.2 = 1 (as expected).

Probability distribution is following

P=0 P=1 P=2

probability 0.3 0.5 0.2