Answer :

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 ^{20}C_{2}

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 ^{16}C_{2} 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 ^{4}C_{1} and number of ways of selecting one good orange out of 16 is ^{16}C_{1}

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 ^{4}C_{2}

Hence

Represent P(X),X in tabular form and calculating XP(X) and X^{2}P(X) for mean and variance

Now mean is given by

⇒ E(X) = 0.4

Now variance is V(X) = E(X^{2}) – [E(X)]^{2}

E(X^{2}) is given by

Hence

Cancel out 5

Hence mean is 0.4 and variance is

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