For each j,1≤j≤12, an element zj is chosen from V at random, independently of the other choices. Let P=∏j=112zj be the product of the 12 numbers selected. What is the probability that P=−1?
Answer Choices:
A. 3105⋅11
B. 2⋅31052⋅11
C. 395⋅11
D. 2⋅3105⋅7⋅11
E. 31022⋅5⋅11 Solution:
If zj is an element of the set A={2i,−2i}, then ∣zj∣=2. Otherwise zj is an element of
B=V\A={81(1+i),81(−1+i),81(1−i),81(−1−i)}
and ∣zj∣=21. It follows that ∣P∣=∏j=112∣zj∣=1 exactly when 8 of the 12 factors zj are in A and 4 of the factors are in B. The product of 8 complex numbers each of which is in A is a real number, either 16 or -16 . The product of 4 numbers each of which is in B is one of 161,161i,−161, or −161i. Thus a product P=∏j=112zj is -1 exactly when 8 of the zj are from A,4 of the zj are from B, and the last of the 4 elements from B is chosen so that the product is -1 rather than i,−i, or 1 . Because the probability is 31 that a particular factor zj is from A, the probability is 32 that a particular factor zj is from B, and the probability is 61 that a particular factor zj is a specific element of V, the probability that the product P will be -1 is given by