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
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