Probability explanation please

Bunuel,
please can you help me?

The question about Xavier (X), Yvonne (Y), and Zelda (Z) involves a particular single outcome out of many possible ones. Specifically, it is asking for the probability of (X, Y, Z’). The question is NOT asking “What is the probability that exactly 2 out of the 3 students answer the question correctly?” If it were, then there would be two additional outcomes that would have to be included in the calculation of the final answer. These two additional outcomes would be (X, Y’, Z) and also (X’, Y, Z), in addition to (X, Y, Z’). But since the question is asking only for the probability that particular individuals answered correctly or incorrectly, (X, Y, Z’), then there is only one outcome whose probability must be calculated: P(X) × P(Y) × P(Z’) = ¼ × ½ ×3/8 = 3/64.
The three additional questions that you addressed all deal with multiple outcomes that satisfy the requirement stated in each problem.

The second question, in which one coin was flipped three times, has three outcomes that satisfy the requirement: (H,H,T) or (H,T,H) or (T, H, H).

Your third question is identical to the second question. It doesn’t matter if one flips one coin three times or if one flips three coins at once. The possible outcomes that satisfy the requirement of “exactly two heads” is identical to those listed in the second question: (H,H,T) or (H,T,H) or (T, H, H).

You stated that your mall question is conceptually identical to the first problem, but it is VERY different from the first question about X, Y, and Z. In fact, it's conceptually identical to the second and third questions. Here’s why: In your mall question, you have 6 visitors at the mall. They are unnamed and non-specific. When you state the mall question “What is the probability that exactly 4 of the 6 buy candy?” then we don’t know and we don’t care who did or did not buy candy, as long as 4 of the 6 bought candy. The answer would be (3/10)^4 * (7/10)^2 * 6!/(4!*2!) where the last factor 6!/(4!*2!) = 15 is the number of arrangements 4 out of 6 visitor buying candy. Some of these arrangements include: (YYYYNN) or (YYYNYN) or (YYNNYY), and there are 12 additional outcomes that satisfy the requirement.

But now let’s give the mall visitors specific identities: Ann (A), Bob (B), Cal (C), Doug (D), Emil (E), and Finn (F).

Now let’s ask a question about specific people (which is what we did in question 1). If we ask, “What is the probability that A, B, C, and D bought candy and E and F did not buy candy?” then there would be only ONE outcome: (A, B, C, D, E’, F’), and that is it. This scenario (with specific names and specific outcomes for those named individuals) is identical in concept in the first question, which asked: “What is the probability that X and Y passed and Z did not pass?”.