Answer:
Step-by-step explanation:
1) The vectors might no span R^4. if you want to span R^4, at least 4 vector of the six must be linear indepent. But, consider the following set of vectors
{(1,1,1,1), (2,2,2,2), (3,3,3,3), (4,4,4,4), (5,5,5,5), (6,6,6,6)}. This set has 6 vectors of R4, but they are all linearly dependent, since they are a linear combination of the vector (1,1,1,1). Since we have only one linear indepent vector, we cannot span R4 with this set.
2) Since R4 is a 4-dimensional vector space, by having 6 vectors in an 4-dimensional space, the must necessarily be linearly dependent, since in R4 the maximum number of linearly independent vectors we could have in a set is 4.
3). The statement may be true. Any four subset of these vectors may be a base. To do so, that depends on their linear independence. If they are linearly independence, they are base since they are 4 linearly independent vectors in a 4-dimensional space, but if they lack linearly independence, they cannot be a base.
For example, consider the following set of vectors
{(1,0,0,0),(0,1,0,0),(0,0,1,0), (0,0,0,1), (1,1,1,1), (0,0,0,1)}.
Note that if we take the first 4 vectors, they are linearly indepent, hence they are a base. Same happens if we take vectors 2-5. But if we take vectors 3-6, then this subset is not a base, since it has only 3 linearly independent vectors.
and 
, respectively. The total probability of knocking down at most two is thus 
, or 
.
