= Solution
Random <sampling without replacement> assigns each of the ten distinguishable pieces to each lunch with the same <probability>. Five pieces of each type therefore give
$$
\boxed{P(A_1)=\frac12,\qquad P(B_2)=\frac12.}
$$
One can check the second <probability> by conditioning on the first lunch:
$$
P(B_2)=P(B_2\mid A_1)P(A_1)+P(B_2\mid B_1)P(B_1)
=\frac59\frac12+\frac49\frac12=\frac12.
$$
The identical marginal <probabilities> do not imply <independence> of the draws.
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