Question 1178260
Let X = random variable equal to the number of remaining red balls.

Case 1. X = 0.
X = 0 means all 7 balls had been drawn, with black as the 7th and last ball. The number of ways of arranging the 6 previous places will be {{{6!/(3!*3!) = 20}}}.

Case 2. X = 1.
X = 1 means 6 balls had been drawn, with black as the 6th ball (and red as the remaining ball.) The number of ways of arranging the 5 previous places will be {{{5!/(3!*2!) = 10}}}.

Case 3. X = 2.
X = 2 means 5 balls had been drawn, with black as the 5th ball (and 2 red balls as the remaining ones.) The number of ways of arranging the 4 previous places will be {{{4!/(3!*1!) = 4}}}

Case 4. X = 3.
X = 3 means 4 balls had been drawn, with black as the 4th ball.  There is only 1 possible way of getting this scenario, 
and that is when all four black balls were drawn one after the other at the very start.


The probability mass function is then as follows:

P(X=0) = 20/35
P(X=1) = 10/35
P(X=2) = 4/35
P(X=3) = 1/35


The expected value of X is then {{{E(X) = 0*(20/35) + 1*(10/35) + 2*(4/35) + 3*(1/35) = 21/35 = 3/5 = highlight(0.6)}}}