Step-by-step 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
