Given \(A = \{1, 2, 3, 4\}\) and \(B = \{3, 4, 5, 6\}\), find \(A \cup B\).
Explanation
Answer: B. \(A \cup B\) combines every element from both sets — remove duplicates. \(\{1,2,3,4\} \cup \{3,4,5,6\} = \{1,2,3,4,5,6\}\). The elements 3 and 4 appear in both, but we list them only once.Using the same sets \(A = \{1,2,3,4\}\) and \(B = \{3,4,5,6\}\), what is \(A \cap B\)?
Careful — this is NOT the same as union!
Explanation
Answer: C. Intersection \(\cap\) = elements that appear in both sets. Only 3 and 4 are in both A and B. Think: the ∩ symbol looks like a bridge connecting two sets — only what's on both sides counts.If \(\mathcal{U} = \{1,2,3,4,5,6,7,8\}\) and \(A = \{2,4,6,8\}\), find \(A'\) (the complement of A).
Explanation
Answer: C. The complement \(A'\) contains all elements in the universal set \(\mathcal{U}\) that are not in A. Remove \(\{2,4,6,8\}\) from \(\{1,2,3,4,5,6,7,8\}\) → remaining = \(\{1,3,5,7\}\). Remember: \(A \cup A' = \mathcal{U}\) always.In a class of 30 students, 18 study French, 15 study Spanish, and 8 study both. How many study at least one language?
Use the inclusion-exclusion principle. Don't double-count!
Explanation
Answer: B. \(n(F \cup S) = 18 + 15 - 8 = 25\). We subtract 8 because those students were counted in both the French group and the Spanish group. Without subtracting, we'd count them twice.Which of the following is true?
Explanation
Answer: B. \(\{3\} \subseteq \{1,2,3,4\}\) is true because 3 is indeed inside \(\{1,2,3,4\}\). For A: 1 is not in \(\{2,3,4\}\). For C: 5 is not in the set. For D: 1 is not in \(\{2,3\}\). A subset means every element of the left set must appear in the right set.In a Venn diagram, region \((A \cap B)'\) represents:
Explanation
Answer: C. The prime ( ′ ) means "complement" = everything outside. \((A \cap B)'\) is the complement of the intersection, meaning all elements NOT in both A and B simultaneously. This includes: only-A elements, only-B elements, and elements outside both circles.A bag contains 4 red, 3 blue, and 5 green marbles. What is the probability of picking a red marble?
Explanation
Answer: B and C are both correct. Total = 4+3+5 = 12. \(P(\text{red}) = \frac{4}{12} = \frac{1}{3}\). Both B and C equal \(\frac{1}{3}\). In this question, C is the unsimplified form. Always simplify your final answer unless told otherwise.The probability it rains tomorrow is 0.35. What is the probability it does NOT rain?
Complementary events always sum to 1.
Explanation
Answer: C. \(P(\text{no rain}) = 1 - P(\text{rain}) = 1 - 0.35 = 0.65\). The key rule: complementary events cover all possibilities, so they always add to exactly 1. D is impossible since probability can never exceed 1.A coin is flipped and a die is rolled. What is the probability of getting a Head AND a 6?
Explanation
Answer: C. \(P(H) = \frac{1}{2},\ P(6) = \frac{1}{6}\). Since the coin and die are independent: \(P(H \cap 6) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}\).A card is drawn from a standard deck of 52. What is the probability it is a King OR a Queen?
A card cannot be both a King and a Queen at the same time — these are mutually exclusive.
Explanation
Answer: C. 4 Kings + 4 Queens = 8 favorable cards out of 52. \(P = \frac{8}{52} = \frac{2}{13}\). Since a card can't be both simultaneously (mutually exclusive), we simply add: \(\frac{4}{52} + \frac{4}{52} = \frac{8}{52}\).In a class, 40% of students play football, 30% play basketball, and 15% play both. Given a student plays football, what is the probability they also play basketball?
This is conditional probability. The "|" symbol means "given that."
Explanation
Answer: B. \(P(B|F) = \frac{0.15}{0.40} = 0.375\). We narrow our sample space to only football players (40%). Among those, 15% also play basketball. So: \(\frac{15}{40} = \frac{3}{8} = 0.375\).A bag has 5 red and 3 blue balls. Two balls are drawn without replacement. What is \(P(\text{both red})\)?
Explanation
Answer: C. 1st draw: \(\frac{5}{8}\). After removing one red, 2nd draw: \(\frac{4}{7}\). Multiply: \(\frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}\). B also equals \(\frac{5}{14}\) — same answer, different form. Without replacement means the total decreases for each draw!Evaluate \(5!\)
Factorial = multiply all positive integers down to 1. Used in permutations & combinations.
Explanation
Answer: C. \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). Memorize common factorials: \(3!=6,\ 4!=24,\ 5!=120,\ 6!=720\). Note: \(0! = 1\) by definition!How many ways can 3 students be chosen and arranged in a line from a group of 8?
Order matters here — 1st, 2nd, 3rd place are different!
Explanation
Answer: B. \(P(8,3) = \frac{8!}{(8-3)!} = \frac{8!}{5!} = 8 \times 7 \times 6 = 336\). Shortcut: start at 8 and multiply downward for 3 factors: \(8 \times 7 \times 6 = 336\).How many ways can a committee of 3 be selected from 8 people?
A committee has no rank — selecting Alice, Bob, Carol is the same as Carol, Bob, Alice.
Explanation
Answer: B. \(\binom{8}{3} = \frac{8!}{3!\,5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56\). Notice: \(C(8,3) = \frac{P(8,3)}{3!} = \frac{336}{6} = 56\). We divide by \(r!\) to remove the counted arrangements of the same group.Which of the following equals \(\dbinom{10}{7}\)?
Hint: use the symmetry property of combinations.
Explanation
Answer: B. \(\binom{n}{r} = \binom{n}{n-r}\). So \(\binom{10}{7} = \binom{10}{10-7} = \binom{10}{3}\). Both equal 120. This symmetry makes sense: choosing 7 to include is the same as choosing 3 to exclude.A restaurant offers 4 starters, 5 mains, and 3 desserts. How many different 3-course meals are possible?
Explanation
Answer: B. Fundamental Counting Principle: multiply. \(4 \times 5 \times 3 = 60\). Each choice is independent, so we multiply all possibilities together.A box has 6 good items and 4 defective items. 2 items are selected at random. What is the probability that both are good?
Explanation
Answer: A. \(\binom{6}{2} = 15\), \(\binom{10}{2} = 45\). \(P = \frac{15}{45} = \frac{1}{3}\). Note: C also equals \(\frac{1}{3}\) — same value, just not simplified. Always reduce to lowest terms.Two dice are rolled. What is the probability that the sum equals 7?
Think systematically — list all pairs that sum to 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1).
Explanation
Answer: A. Total outcomes = \(6 \times 6 = 36\). Favorable outcomes (sum = 7): (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) = 6 outcomes. \(P = \frac{6}{36} = \frac{1}{6}\). Sum of 7 is actually the most likely sum on two dice!A password consists of 2 letters (A–Z, no repeat) followed by 3 digits (0–9, no repeat). How many passwords are possible?
Order matters for passwords! Use permutations.
Explanation
Answer: A. Letters: first slot = 26 choices, second = 25 (no repeat). Digits: first = 10, second = 9, third = 8. Total = \(26 \times 25 \times 10 \times 9 \times 8 = 468{,}000\). B is wrong because combinations don't account for order — "AB" and "BA" are different passwords. C ignores the "no repeat" rule.Your Final Score
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