Question

From a box of fruit containing 82 oranges and 1 apple a random sample of 2 pieces of fruit has been selected without replacement. Let X be the number of oranges and Y be the number of apples in the sample. What will the Variance of Y be?

Answer 1

Let Y be the number of apples in the sample.

Sample space of Y is 0 and 1.

Probability distribution of X is as follows

Y	Y=0	Y =1
P(Y=y)	no apple drawn that is both fruits are oranges = (82/83) ×( 81/82) = 81/83	One apple is drawn (2 possible case: once in 1st draw and second in 2nd draw) = (1/83) × ( 82/82) + (82/83)×(1/82) = 2/83

We have,

E(Y) = summation ( Y × P(Y = y))

= 0 × 81/83 + 1 × 2/83 = 2/83

and E(Y2 ) = summation ( Y2 × P(Y = y))

= 0×0 × 81/83 + 1×1 × 2/83 = 2/83

So, by definition,

Variance of Y = E(Y2 ) - (E(Y)) 2 = (2/83) - (2/83)2

= 162 / 6889 = 0.0235