Question

An urn contains three red balls and four blue balls. Draw two balls at random from the urn, without replacement. Compute the expected number of red balls in your sample. (Round your answer to four decimal places.)

Answer 1

Explanation:

in total we have 3+4 = 7 balls.

when we draw the first ball, the probability to draw a red ball is 3/7, and a blue ball 4/7.

when we draw the second ball, we have now only 6 balls in total.

the probabilty to draw a red back now depends also on the result of the first draw.

if the first ball was already red, then we have only a chance now of 2 out of 6.

if the first ball was blue, then we have now a chance of 3 out of 6.

so, the probably to draw at least 1 red ball in 2 draws is the probability of drawing one on the first draw plus the probability of drawing one on the second :

1 - probability to see 2 blue balls

1 - 4/7 × 3/6 = 1 - 12/42 = 30/42 = 0.714285714...

the expected number of red balls in 2 draws is

1 red in first red in first red in second

but not second and second but not first

1×(3/7 × 4/6) + 2×(3/7 × 2/6) + 1×(4/7 × 3/6) = 12/42 + 12/42 + 12/42 = 36/42 = 6/7 = 0.857142857 ≈ 0.8571