Four bad oranges are accidentally mixed with 16 good ones. Find the probability distribution of the number of bad oranges when two oranges are drawn at random from this lot. Find the mean and variance of the distribution.

4 bad oranges get mixed with 16 hence total oranges are 20.

Now 2 oranges are selected at random from 20 oranges.

Number of ways in which 2 oranges can be selected out of 20 is ²⁰C₂

We have to find the probability distribution of number of bad oranges let X be the variable which will represent it.

Now out of the 2 oranges selected none can be bad or 1 can be bad or both.

Hence X can take values 0, 1 and 2

Now let us find the probabilities P (X = 0), P (X = 1) and P(X = 2)

P (X = 0) means probability of getting 0 bad oranges that is both the oranges selected are good

Number of ways in which 2 good oranges can be selected is ¹⁶C₂ because there are total 16 good oranges out of which 2 are chosen

Hence

P (X = 1) means probability of selecting one bad orange that is one good and one bad orange

Number of ways in which 1 bad orange can be selected out of 4 is ⁴C₁ and number of ways of selecting one good orange out of 16 is ¹⁶C₁

Hence

P(X = 2) means selecting both the bad oranges

Number of ways in which 2 bad oranges can be selected out of 4 is ⁴C₂

Hence

Represent P(X),X in tabular form and calculating XP(X) and X²P(X) for mean and variance

Now mean is given by

⇒ E(X) = 0.4

Now variance is V(X) = E(X²) – [E(X)]²

E(X²) is given by

Hence

Cancel out 5

Hence mean is 0.4 and variance is