Is x + x^3 one term
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Without replacement from a small (8) number of balls means the binomial distribution is NOT applicable since the probability of each pick is not constant.
We can either use the hypergeometric distribution, or solve from first principles.
For simplicity for the given problem, we will solve it using the hypergeometric distribution.
HypergeometricUse hypergeometric distribution where:a=number of target items selectedA=total number of target items in the "pool" (3 white balls)b=number of non-target items selected (B=total number of non-target items in the "pool" (5 blue balls)Then
where 
= combination of r items selected from n,A+B=total number of items = 3+5 = 8a+b=number of items selected=4
Case 1 : 2 white balls, => a=2, b=2, A=3, B=5
P(2W)=C(A,a)*C(B,b)/C(A+B, a+b)
=C(3,2)*C(5,2)/C(8,4)
=3*10/70
=3/7
Case 2: 3 white balls => a=3, b=1, A=3, B=5
P(3W)=C(3,3)*C(5,1)/C(8,4)
=1*5/70
=1/14
Therefore , by the rule of addition, the probability that two or three of the balls are white
P(2W ∪ 3W) = 3/7 + 1/14 = (6+1)/14 = 7/14 = 1/2
We can either use the hypergeometric distribution, or solve from first principles.
For simplicity for the given problem, we will solve it using the hypergeometric distribution.
HypergeometricUse hypergeometric distribution where:a=number of target items selectedA=total number of target items in the "pool" (3 white balls)b=number of non-target items selected (B=total number of non-target items in the "pool" (5 blue balls)Then
Case 1 : 2 white balls, => a=2, b=2, A=3, B=5
P(2W)=C(A,a)*C(B,b)/C(A+B, a+b)
=C(3,2)*C(5,2)/C(8,4)
=3*10/70
=3/7
Case 2: 3 white balls => a=3, b=1, A=3, B=5
P(3W)=C(3,3)*C(5,1)/C(8,4)
=1*5/70
=1/14
Therefore , by the rule of addition, the probability that two or three of the balls are white
P(2W ∪ 3W) = 3/7 + 1/14 = (6+1)/14 = 7/14 = 1/2