This problem requires counting the ways to select three marbles from ten (six red, four blue) without replacement to find the probability of exactly two red. Since only the colors matter and not the order drawn, order does not matter—selecting red-blue-red is the same as blue-red-red. The total possible outcomes are the combinations of choosing 3 out of 10, which is $ C(10,3) = 120 $. The favorable outcomes are choosing exactly 2 red from 6 and 1 blue from 4, which is $ C(6,2) \times C(4,1) = 15 \times 4 = 60 $. Thus, the probability is $ [C(6,2) \times C(4,1)] / C(10,3) $. A common mistake is using permutations when order is irrelevant, overcounting; always ask, 'Would switching the order of drawn marbles create a new outcome?' Here, no, since we care only about the set, so use combinations.

This problem requires counting the podium outcomes for seven runners awarding distinct medals to find the probability Lina gets a medal. Since gold, silver, bronze are different, order matters—first place differs from second. The total possible outcomes are the permutations of assigning top 3 out of 7, which is $7P3 = 7 \times 6 \times 5$. The favorable outcomes place Lina in one of the three spots (3 choices), then assign the other two from the remaining 6, which is $3 \times 6P2$. Thus, the probability is $(3 \times 6P2) / 7P3$. A common mistake is treating finishes as unordered sets; always ask, 'Would switching two runners' medals create a new outcome?' Here, yes, because gold and silver are distinct, so use permutations.

This problem requires counting the possible ways to assign three distinct officer positions from eight club members to find the probability that Alex is selected for one. Since the positions are different, order matters—assigning someone as President differs from Vice President. The total possible outcomes are the permutations of selecting and assigning 3 members out of 8, which is 8P3 = 8 × 7 × 6. The favorable outcomes are those where Alex is assigned to one of the three positions: there are 3 choices for Alex's position, and then 7P2 ways to assign the remaining two positions from the other 7 members. Thus, the probability is (3 × 7P2) / 8P3. A common mistake is using combinations instead, like treating selections as unordered, but this ignores the distinct roles; always ask, 'Would switching two people's positions create a new outcome?' Here, yes, because positions are unique, so permutations are appropriate.

This problem requires counting the ways to form four-letter passwords from seven distinct letters (no repeats) to find the probability it starts with A. Since sequences matter, order matters—ABCD differs from ACBD. The total possible outcomes are the permutations of choosing and arranging 4 out of 7, which is 7P4 = 7 × 6 × 5 × 4. The favorable outcomes fix A first, then choose and arrange 3 from the remaining 6, which is 6P3 = 6 × 5 × 4. Thus, the probability is 6P3 / 7P4. A common mistake is using combinations, treating order as irrelevant; always ask, 'Would switching two letters' positions create a new outcome?' Here, yes, like swapping first and second changes the password, so permutations are needed.

This problem requires counting the ways to choose a three-person committee from seven (four seniors, three juniors) to find the probability of exactly two seniors and one junior. Since no roles are assigned, order does not matter—the committee is just a group. The total possible outcomes are the combinations of choosing 3 out of 7, which is C(7,3) = 35. The favorable outcomes are choosing 2 seniors from 4 and 1 junior from 3, which is C(4,2) × C(3,1) = 6 × 3 = 18. Thus, the probability is [C(4,2) × C(3,1)] / C(7,3). A common mistake is using permutations for ordered selection when it's unnecessary; always ask, 'Would switching two members' positions create a new outcome?' Here, no, since there are no positions, so combinations work best.

This problem requires counting the ways to select two socks from eight (five black, three white) without replacement to find the probability they match in color. Since only colors matter, order does not matter—selecting black then white is the same as white then black. The total possible outcomes are the combinations of choosing 2 out of 8, which is C(8,2) = 28. The favorable outcomes are both black, C(5,2) = 10, or both white, C(3,2) = 3, totaling 13. Thus, the probability is [C(5,2) + C(3,2)] / C(8,2). A common mistake is using permutations when order is irrelevant, like counting ordered pairs; always ask, 'Would switching the order of socks create a new outcome?' Here, no, since we care only about the pair, so use combinations.

This problem requires counting the ways to choose a four-person committee from ten students to find the probability that Maya and Noah are both included. Since no roles are assigned, order does not matter— the committee is just a set of people. The total possible outcomes are the combinations of choosing 4 out of 10, which is C(10,4) = 210. The favorable outcomes fix Maya and Noah in the committee, then choose 2 more from the remaining 8, which is C(8,2) = 28. Thus, the probability is C(8,2) / C(10,4). A common mistake is treating it as ordered when it's not, like using permutations; always ask, 'Would switching two members' positions create a new outcome?' Here, no, since there are no positions, so combinations are correct.

This problem requires counting the ways to award three distinct prizes to nine entrants to find the probability that Jordan wins any prize. Since prizes are different (1st, 2nd, 3rd), order matters—assigning 1st to someone differs from 2nd. The total possible outcomes are the permutations of selecting and assigning 3 out of 9, which is $9P3 = 9 \times 8 \times 7$. The favorable outcomes assign Jordan to one of the three prizes (3 choices), then assign the remaining two prizes from the other 8, which is $3 \times 8P2$. Thus, the probability is $(3 \times 8P2) / 9P3$. A common mistake is ignoring the distinct prizes and using combinations; always ask, 'Would switching two winners' prizes create a new outcome?' Here, yes, because 1st and 2nd are unique, so use permutations.

This problem requires counting committee selections to find the probability a specific candidate is included. Since order does not matter, we use combinations, as committee membership is about the set, not arrangement. The total possible outcomes are choosing 3 from 7, which is $\binom{7}{3}$. The favorable outcomes include Jordan and choose 2 more from the remaining 6, which is $\binom{6}{2}$. Thus, the probability is $\dfrac{\binom{6}{2}}{\binom{7}{3}}$. A common mistake is using permutations when committees are unordered, overcounting by considering roles. To decide, ask: 'Would switching two members create a new committee?' If no, as here, use combinations.

This problem requires counting student selections to find the probability both are girls. Since order does not matter in the selection, we use combinations, as the pair is what counts, not who is chosen first. The total possible outcomes are choosing any 2 from 9, which is (\binom{9}{2}). The favorable outcomes are choosing 2 girls from 4, which is (\binom{4}{2}). Thus, the probability is (\d$\frac{\binom{4}{2}$}{\binom{9}{2}}). A common mistake is using permutations for unordered selections, like treating the representatives as ordered. To check, ask: 'Would switching the two selected students create a new outcome?' If no, as in this group, use combinations.