Statistics Quiz: Using Permutations And Combinations In Probability

What this quiz covers

This quiz focuses on Using Permutations And Combinations In Probability, giving you a quick way to practice the rules, question types, and explanations that matter most for Statistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\dfrac{{6\choose 2}}{{7\choose 3}}$
2. $\dfrac{3\cdot 6P2}{7P3}$ (correct answer)
3. $\dfrac{3}{7}$
4. $\dfrac{6P2}{7P3}$

Explanation: 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.

Question 2

1. $\dfrac{1}{7P4}$
2. $\dfrac{6C3}{7C4}$
3. $\dfrac{6P3}{7P4}$ (correct answer)
4. $\dfrac{7P3}{7P4}$

Explanation: 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.

Question 3

1. $\dfrac{4P2\cdot 3P1}{7P3}$
2. $\dfrac{{4\choose 2}{3\choose 1}}{{7\choose 3}}$ (correct answer)
3. $\dfrac{{4\choose 2}{3\choose 1}}{7P3}$
4. $\dfrac{{4\choose 2}{3\choose 1}}{{7\choose 2}}$

Explanation: 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.

Question 4

1. $\dfrac{{5\choose 2}+{3\choose 2}}{{8\choose 2}}$ (correct answer)
2. $\dfrac{{5\choose 2}+{3\choose 2}}{8^2}$
3. $\dfrac{{5\choose 1}{3\choose 1}}{{8\choose 2}}$
4. $\dfrac{5P2+3P2}{8P2}$

Explanation: 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.

Question 5

1. $\dfrac{5!}{6!}$
2. $\dfrac{2\cdot 4!}{6!}$
3. $\dfrac{2\cdot 5!}{6!}$ (correct answer)
4. $\dfrac{2}{6!}$

Explanation: This problem requires counting the arrangements of six distinct photos in a row to find the probability that Photo X is at one end. Since it's a left-to-right display, order matters—swapping positions changes the arrangement. The total possible outcomes are all permutations of the six photos, which is 6! = 720. The favorable outcomes place X at the left end (then arrange the other 5: 5!) or right end (another 5!), totaling 2 × 5! = 240. Thus, the probability is (2 × 5!) / 6!. A common mistake is undercounting by forgetting both ends; always ask, 'Would switching Photo X to the other end create a new outcome?' Here, yes, left and right are distinct, so include both cases separately.

Question 6

1. $\dfrac{3\cdot 8P2}{9P3}$ (correct answer)
2. $\dfrac{8P2}{9P3}$
3. $\dfrac{1}{9}$
4. $\dfrac{{8\choose 2}}{{9\choose 3}}$

Explanation: 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.

Question 7

1. $\dfrac{1}{6}$ (correct answer)
2. $\dfrac{5\mathrm{P}3}{6\mathrm{P}4}$
3. $\dfrac{4}{6}$
4. $\dfrac{6\mathrm{P}4}{5\mathrm{P}3}$

Explanation: This problem requires counting digit arrangements to find the probability the code starts with 1. Since it's a code and order matters, we use permutations, as different sequences are unique codes. The total possible outcomes are arranging 4 distinct digits from 6, which is $6 \times 5 \times 4 \times 3$. The favorable outcomes fix 1 first, then arrange 3 from the remaining 5, which is $1 \times 5 \times 4 \times 3$. Thus, the probability simplifies to $\dfrac{1}{6}$. A common mistake is using combinations for ordered codes, undercounting distinct arrangements. To avoid this, ask: 'Would switching two digits create a new code?' If yes, like here, use permutations.

Question 8

1. $\dfrac{1}{6\cdot 5\cdot 4}$
2. $\dfrac{1}{\binom{6}{3}}$
3. $\dfrac{1}{6\cdot 5}$ (correct answer)
4. $\dfrac{4}{6\cdot 5\cdot 4}$

Explanation: This problem requires counting medal assignments to find the probability of specific winners for gold and silver. Since medals are distinct and order matters, we use permutations, as who gets which medal changes the outcome. The total possible outcomes are assigning 3 medals from 6 runners, which is (6 \times 5 \times 4). The favorable outcomes have Alex gold (1 way), Blake silver (1 way), and any of 4 for bronze. Thus, the probability simplifies to (\dfrac{1}{6 \times 5}). A common mistake is using combinations when orders are distinct, like treating medals as identical. To avoid confusion, ask: 'Would switching gold and silver create a new outcome?' If yes, as in this race, use permutations.

Question 9

1. $\dfrac{2}{4}$
2. $\dfrac{8}{24}$
3. $\dfrac{6}{24}$
4. $\dfrac{12}{24}$ (correct answer)

Explanation: This problem requires counting seating arrangements to find the probability that two specific students are adjacent. Since seats are in a row and order matters, we use permutations, as swapping positions creates a different seating. The total possible outcomes are the arrangements of 4 students, which is (4! = 24). The favorable outcomes treat A and B as a single block (in AB or BA order), leaving 3 units to arrange in (3! \times 2 = 12) ways. Thus, the probability is (\dfrac{12}{24}). A common mistake is ignoring that order matters and using combinations, undercounting distinct seatings. To decide, ask: 'Would switching A and B's positions create a new outcome?' If yes, like here where seats are distinct, use permutations.

Question 10

1. $\dfrac{\binom{6}{3}}{\binom{10}{3}}$
2. $\dfrac{\binom{6}{2}\binom{4}{2}}{\binom{10}{3}}$
3. $\dfrac{\binom{6}{2}\binom{4}{1}}{10\cdot 9\cdot 8}$
4. $\dfrac{\binom{6}{2}\binom{4}{1}}{\binom{10}{3}}$ (correct answer)

Explanation: This problem requires counting marble selections to find the probability of a specific color mix. Since order does not matter in the draw, we use combinations, as the set of marbles is what counts, not the sequence. The total possible outcomes are choosing any 3 marbles from 10, which is (\binom{10}{3}). The favorable outcomes are selecting exactly 2 red from 6 and 1 blue from 4, given by (\binom{6}{2} \times \binom{4}{1}). Thus, the probability is (\dfrac{\binom{6}{2}\binom{4}{1}}{\binom{10}{3}}). A common mistake is using permutations for unordered draws, overcounting by considering draw sequences. To avoid confusion, ask: 'Would switching the order of drawn marbles create a new outcome?' If no, like here, stick to combinations.

Question 11

1. $\dfrac{\binom{5}{2}\binom{3}{1}}{\binom{8}{2}}$
2. $\dfrac{\binom{5}{2}\binom{3}{1}}{\binom{8}{3}}$ (correct answer)
3. $\dfrac{\binom{5}{3}}{\binom{8}{3}}$
4. $\dfrac{\binom{5}{2}\binom{3}{1}}{8\cdot 7\cdot 6}$

Explanation: This problem requires counting the number of ways to choose a committee to find the probability of a specific composition. Since it's a committee and order does not matter, we use combinations, as swapping members doesn't create a new group. The total possible outcomes are the ways to choose any 3 students from 8, which is (\binom{8}{3}). The favorable outcomes are choosing exactly 2 seniors from 5 and 1 junior from 3, calculated as (\binom{5}{2} \times \binom{3}{1}). Thus, the probability is (\dfrac{\binom{5}{2}\binom{3}{1}}{\binom{8}{3}}). A common mistake is using permutations when order doesn't matter, like treating committees as ordered, which overcounts. To avoid this, ask: 'Would switching two members create a new committee?' If no, use combinations, as in this example where order is irrelevant.

Question 12

1. $\dfrac{\binom{3}{1}\binom{5}{3}}{\binom{8}{4}}$
2. $\dfrac{\binom{5}{1}}{\binom{8}{4}}$
3. $\dfrac{\binom{3}{3}\binom{5}{1}}{\binom{8}{4}}$ (correct answer)
4. $\dfrac{\binom{3}{3}\binom{5}{1}}{8\cdot 7\cdot 6\cdot 5}$

Explanation: This problem requires counting candy selections to find the probability of drawing all green plus one yellow. Since order does not matter, we use combinations, as the handful is a set, not a sequence. The total possible outcomes are choosing 4 from 8, which is (\binom{8}{4}). The favorable outcomes are all 3 green from 3 and 1 yellow from 5, given by (\binom{3}{3} \times \binom{5}{1}). Thus, the probability is (\dfrac{\binom{3}{3}\binom{5}{1}}{\binom{8}{4}}). A common mistake is using permutations for unordered draws, inflating counts with sequences. To check, ask: 'Would switching two candies in the draw create a new outcome?' If no, like this jar draw, use combinations.

Question 13

1. $\dfrac{\binom{7}{2}}{\binom{8}{3}}$
2. $\dfrac{1}{8}$
3. $\dfrac{1}{8\mathrm{P}3}$
4. $\dfrac{7\mathrm{P}2}{8\mathrm{P}3}$ (correct answer)

Explanation: This problem asks for the probability that Jordan wins exactly 2nd place in a contest with ordered places. Since the places (1st, 2nd, 3rd) are distinct, order matters. The total possible outcomes are 8P3 = 8×7×6 = 336 ways to assign three different places to 8 finalists. For favorable outcomes (Jordan in 2nd place), we fix Jordan in 2nd position and need to fill 1st and 3rd places with 2 of the remaining 7 finalists, giving 7P2 = 7×6 = 42 ways. Therefore, the probability is 7P2/8P3 = 42/336 = 1/8. A common mistake is overthinking this - once Jordan is fixed in 2nd place, we simply arrange the others in the remaining positions using permutations.

Question 14

1. $1-\dfrac{\binom{7}{3}}{\binom{12}{3}}$ (correct answer)
2. $1-\dfrac{7\mathrm{P}3}{12\mathrm{P}3}$
3. $\dfrac{\binom{7}{3}}{\binom{12}{3}}$
4. $\dfrac{\binom{5}{1}\binom{7}{2}}{\binom{12}{3}}$

Explanation: This problem asks for the probability that at least 1 left-handed student is selected. Since it's a group selection, order doesn't matter. The total possible outcomes are C(12,3) = 220 ways to choose 3 students from 12. For "at least 1 left-handed," it's easier to use the complement: find the probability of NO left-handed students and subtract from 1. The number of ways to choose 3 students from the 7 right-handed students is C(7,3) = 35. So the probability of no left-handed students is C(7,3)/C(12,3) = 35/220 = 7/44. Therefore, the probability of at least 1 left-handed student is 1 - 7/44 = 37/44. When you see "at least one," consider using the complement approach: it's often simpler than counting multiple cases.

Question 15

1. $\dfrac{\binom{6}{3}}{\binom{2}{1}\binom{2}{1}\binom{2}{1}}$
2. $\dfrac{2\mathrm{P}1\cdot 2\mathrm{P}1\cdot 2\mathrm{P}1}{6\mathrm{P}3}$
3. $\dfrac{\binom{2}{1}\binom{2}{1}\binom{2}{1}}{6\mathrm{P}3}$
4. $\dfrac{\binom{2}{1}\binom{2}{1}\binom{2}{1}}{\binom{6}{3}}$ (correct answer)

Explanation: This problem asks for the probability of drawing one sock of each color. Since we only care about which socks are drawn (not the order), we use combinations. The total possible outcomes are C(6,3) = 20 ways to choose any 3 socks from 6. For favorable outcomes (one red, one blue, one green), we need C(2,1) ways to choose 1 red from 2 red socks AND C(2,1) ways to choose 1 blue from 2 blue socks AND C(2,1) ways to choose 1 green from 2 green socks. This gives C(2,1)×C(2,1)×C(2,1) = 2×2×2 = 8 ways. Therefore, the probability is [C(2,1)×C(2,1)×C(2,1)]/C(6,3) = 8/20 = 2/5. Remember: drawing {red₁, blue₂, green₁} gives the same set as {blue₂, green₁, red₁}, so order doesn't matter.

Question 16

1. $\dfrac{\binom{5}{2}\binom{3}{1}}{\binom{8}{2}}$
2. $\dfrac{\binom{5}{3}}{\binom{8}{3}}$
3. $\dfrac{5\mathrm{P}2\cdot 3\mathrm{P}1}{8\mathrm{P}3}$
4. $\dfrac{\binom{5}{2}\binom{3}{1}}{\binom{8}{3}}$ (correct answer)

Explanation: This problem asks about drawing marbles where only the set of marbles matters, not the order drawn. Since drawing {red, red, blue} gives the same set as {red, blue, red}, order doesn't matter and we use combinations. The total possible outcomes are C(8,3) = 56 ways to choose any 3 marbles from 8. For favorable outcomes (exactly 2 red and 1 blue), we need C(5,2) ways to choose 2 red from 5 AND C(3,1) ways to choose 1 blue from 3, giving C(5,2) × C(3,1) = 10 × 3 = 30 ways. Therefore, the probability is [C(5,2) × C(3,1)]/C(8,3) = 30/56 = 15/28. Remember: when drawing marbles into a group, rearranging them doesn't create a new outcome, so use combinations.

Question 17

1. $\dfrac{5\mathrm{P}3\cdot 4\mathrm{P}1+5\mathrm{P}4\cdot 4\mathrm{P}0}{9\mathrm{P}4}$
2. $\dfrac{\binom{5}{3}\binom{4}{1}}{\binom{9}{4}}$
3. $\dfrac{\binom{5}{3}\binom{4}{1}+\binom{5}{4}\binom{4}{0}}{9\mathrm{P}4}$
4. $\dfrac{\binom{5}{3}\binom{4}{1}+\binom{5}{4}\binom{4}{0}}{\binom{9}{4}}$ (correct answer)

Explanation: This problem requires finding the probability of selecting a team with at least 3 experienced students. Since it's a team selection, order doesn't matter. The total possible outcomes are C(9,4) = 126 ways to choose 4 students from 9. For favorable outcomes, "at least 3 experienced" means exactly 3 experienced OR exactly 4 experienced students. This gives C(5,3)×C(4,1) + C(5,4)×C(4,0) = 10×4 + 5×1 = 40 + 5 = 45 ways. Therefore, the probability is [C(5,3)×C(4,1) + C(5,4)×C(4,0)]/C(9,4) = 45/126 = 5/14. When you see "at least," remember to add up all qualifying cases. Since rearranging team members doesn't create a new team, use combinations throughout.

Question 18

1. $\dfrac{2\mathrm{P}2\cdot 7\mathrm{P}2\cdot 4!}{9\mathrm{P}4}$
2. $\dfrac{2\mathrm{P}2\cdot 7\mathrm{P}2}{9\mathrm{P}4}$ (correct answer)
3. $\dfrac{2\mathrm{P}2\cdot 7\mathrm{P}2}{9\mathrm{P}3}$
4. $\dfrac{2\mathrm{P}2\cdot 7\mathrm{P}2}{\binom{9}{4}}$

Explanation: This problem involves selecting and arranging books in a row, where order matters because different arrangements create different displays. The total possible outcomes are 9P4 = 9×8×7×6 = 3024 ways to arrange 4 books from 9. For favorable outcomes (both Book X and Book Y included), we first choose positions for X and Y (2P2 = 2 ways to arrange them in 2 of the 4 positions), then fill the remaining 2 positions with 2 of the other 7 books (7P2 = 7×6 = 42 ways). This gives 2P2 × 7P2 = 2 × 42 = 84 ways. Therefore, the probability is (2P2 × 7P2)/9P4 = 84/3024 = 1/36. Ask yourself: "Is X-Y-A-B different from Y-X-A-B?" Yes, so use permutations throughout.

Question 19

1. $\dfrac{\binom{6}{2}\binom{4}{1}}{10\mathrm{P}3}$
2. $\dfrac{\binom{6}{2}\binom{4}{1}}{\binom{10}{3}}$ (correct answer)
3. $\dfrac{\binom{6}{2}\binom{4}{1}}{\binom{10}{2}}$
4. $\dfrac{6\mathrm{P}2\cdot 4\mathrm{P}1}{10\mathrm{P}3}$

Explanation: This problem asks for the probability of selecting a committee with exactly 2 girls and 1 boy. Since it's a committee, order doesn't matter - the committee {Alice, Beth, Charlie} is the same as {Beth, Charlie, Alice}. The total possible outcomes are C(10,3) = 120 ways to choose any 3 students from 10. The favorable outcomes require choosing 2 girls from 6 AND 1 boy from 4, which gives C(6,2) × C(4,1) = 15 × 4 = 60 ways. Therefore, the probability is [C(6,2) × C(4,1)]/C(10,3) = 60/120 = 1/2. A common error is using permutations for committees - ask yourself: "Does rearranging the committee members create a new committee?" No, so use combinations.

Question 20

1. $\dfrac{1}{8}$
2. $\dfrac{1}{56}$ (correct answer)
3. $\dfrac{1}{28}$
4. $\dfrac{1}{336}$

Explanation: This problem involves assigning 8 students to 4 distinct presentation slots, where Lina must be in the 1st slot and Omar in the 4th slot. Since the slots are numbered differently (1st, 2nd, 3rd, 4th), order matters—being assigned to the 1st slot is different from the 2nd slot. The total ways to assign 4 students from 8 to these distinct slots is 8 × 7 × 6 × 5 = 1680 (using permutations). For favorable outcomes where Lina is 1st and Omar is 4th, we have 1 choice for 1st (Lina), 6 choices for 2nd (anyone except Lina and Omar), 5 choices for 3rd, and 1 choice for 4th (Omar), giving us 1 × 6 × 5 × 1 = 30 ways. The probability is 30/1680 = 1/56. Don't confuse this with choosing a group—ask: "Is Lina in slot 1 different from Lina in slot 2?" Yes, because the slots have different positions!