Probability & Complements - Accuplacer Qas

The probability of winning a game on a single try is $p$ . If two independent attempts are made, the probability of winning at least once is $\frac{9}{25}$ . What is the probability of not winning on a single try?
Answer: $\frac{4}{5}$ Let $q$ be the probability of not winning on a single try. Then $P(\text{not winning on either try}) = q^2$ . The event 'winning at least once' is the complement of 'not winning on either try'. Therefore, $P(\text{at least one win}) = 1 - q^2$ . We are given this probability is $\frac{9}{25}$ . So, $\frac{9}{25} = 1 - q^2$ . Solving for $q^2$ gives $q^2 = 1 - \frac{9}{25} = \frac{16}{25}$ . Taking the square root, $q = \frac{4}{5}$ . The probability of not winning on a single try is $\frac{4}{5}$ .
A weather forecast states there is a 25% chance of rain on Saturday and a 30% chance of rain on Sunday. If the weather on the two days is independent, what is the probability that it does not rain on either day?
Answer: 0.525 The probability of no rain on Saturday is $1 - 0.25 = 0.75$ . The probability of no rain on Sunday is $1 - 0.30 = 0.70$ . Since the events are independent, the probability of no rain on either day is the product of these probabilities: $0.75 \times 0.70 = 0.525$ .
An integer is selected at random from the set $\{1, 2, 3, ..., 25\}$ . What is the probability that the selected integer is not a perfect square?
Answer: $\frac{4}{5}$ The perfect squares in the set are $1^2=1, 2^2=4, 3^2=9, 4^2=16, 5^2=25$ . There are 5 perfect squares. The probability of selecting a perfect square is $\frac{5}{25} = \frac{1}{5}$ . The probability of not selecting a perfect square is the complement: $1 - \frac{1}{5} = \frac{4}{5}$ .
A bag contains 6 red marbles, 4 blue marbles, and 5 green marbles. If one marble is drawn at random, what is the probability that the marble is not blue?
Answer: $\frac{11}{15}$ The total number of marbles is $6 + 4 + 5 = 15$ . The probability of drawing a blue marble is $\frac{4}{15}$ . The complement event, drawing a marble that is not blue, has a probability of $1 - P(\text{blue}) = 1 - \frac{4}{15} = \frac{11}{15}$ . Alternatively, the number of non-blue marbles is $6 + 5 = 11$ , so the probability is $\frac{11}{15}$ .
If the probability that an event will occur is $p$ , which expression represents the probability that the event will not occur?
Answer: $1 - p$ By definition, the complement of an event is the set of all outcomes that are not in the event. The sum of the probability of an event and its complement is always 1. If the probability of an event is $p$ , the probability of its complement (the event not occurring) is $1 - p$ .
A student takes a three-question true-or-false quiz and guesses randomly on every question. What is the probability that the student answers at least one question correctly?
Answer: $\frac{7}{8}$ The event 'at least one correct' is the complement of the event 'no questions correct' (all wrong). The probability of guessing a single question incorrectly is $\frac{1}{2}$ . Since the guesses are independent, the probability of guessing all three incorrectly is $(\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) = \frac{1}{8}$ . Therefore, the probability of getting at least one correct is $1 - \frac{1}{8} = \frac{7}{8}$ .
A box contains 50 tickets. A student wins a prize if they draw a ticket numbered with a multiple of 6. What is the probability that a student does not win a prize on a single draw?
Answer: $\frac{21}{25}$ The multiples of 6 up to 50 are 6, 12, 18, 24, 30, 36, 42, 48. There are 8 such numbers. The probability of winning is $\frac{8}{50} = \frac{4}{25}$ . The probability of not winning is the complement: $1 - \frac{4}{25} = \frac{21}{25}$ .
A security system has two independent alarms. The probability that the first alarm fails is 0.1, and the probability that the second alarm fails is 0.2. What is the probability that at least one alarm works correctly?
Answer: 0.98 The event 'at least one alarm works' is the complement of the event 'both alarms fail'. Since the alarms are independent, the probability that both fail is the product of their individual failure probabilities: $P(\text{both fail}) = 0.1 \times 0.2 = 0.02$ . The probability that at least one alarm works is $1 - P(\text{both fail}) = 1 - 0.02 = 0.98$ .
In a quality control process, the probability that a randomly selected item passes inspection is 0.85. If an item fails inspection, it undergoes a repair process that has a 0.70 probability of success. What is the probability that a randomly selected item either passes initial inspection OR is successfully repaired after failing?
Answer: 0.955 This requires finding P(passes initial inspection OR successfully repaired). P(passes) = 0.85. P(fails) = 1 - 0.85 = 0.15. P(successfully repaired | fails) = 0.70, so P(fails AND successfully repaired) = 0.15 × 0.70 = 0.105. Therefore, P(passes OR successfully repaired) = 0.85 + 0.105 = 0.955. Choice B incorrectly multiplies 0.85 × 0.70. Choice C uses the average of the two probabilities. Choice D ignores the repair process entirely.
Events X and Y are independent with P(X) = 0.6 and P(Y) = 0.4. Consider the statement: 'Either X occurs or Y does not occur (or both).' What is the probability that this statement is true?
Answer: 0.84 When you encounter probability questions involving "either...or" statements, you're dealing with union probability. The key is translating the English into mathematical notation and applying the correct probability rules. The statement "Either X occurs or Y does not occur (or both)" translates to $P(X \cup Y^c)$ , where $Y^c$ represents "Y does not occur." Since these events can overlap (both X can occur AND Y can fail to occur), you need the union formula: $P(X \cup Y^c) = P(X) + P(Y^c) - P(X \cap Y^c)$ . First, find $P(Y^c) = 1 - P(Y) = 1 - 0.4 = 0.6$ . Since X and Y are independent, X and $Y^c$ are also independent, so $P(X \cap Y^c) = P(X) \times P(Y^c) = 0.6 \times 0.6 = 0.36$ . Therefore: $P(X \cup Y^c) = 0.6 + 0.6 - 0.36 = 0.84$ . The answer is B. Choice A (0.76) likely comes from incorrectly calculating $P(X) + P(Y^c) - P(Y) = 0.6 + 0.6 - 0.4$ . Choice C (0.60) represents just $P(X)$ , ignoring the "or Y does not occur" part entirely. Choice D (0.64) might result from adding $P(X) + P(Y^c) - P(X)P(Y)$ instead of the correct intersection. Remember: "Either A or B" means union ( $A \cup B$ ), and always subtract the intersection to avoid double-counting when events can occur simultaneously. Practice translating English probability statements into mathematical notation first.
Event A has probability $\frac{2}{5}$ and event B has probability $\frac{3}{8}$ . If events A and B are mutually exclusive, what is the probability of the complement of (A or B)?
Answer: $\frac{9}{40}$ Since A and B are mutually exclusive, P(A or B) = P(A) + P(B) = 2/5 + 3/8 = 16/40 + 15/40 = 31/40. The complement of (A or B) is 1 - P(A or B) = 1 - 31/40 = 9/40. Choice B gives P(A or B) instead of its complement. Choice C incorrectly uses P(A) × P(B) = 6/40. Choice D uses an incorrect calculation method.
A weather forecast states there is a 70% chance of precipitation. Which statement below accurately describes the complement of this event?
Answer: There is a 30% chance of no precipitation. The complement of an event includes all possible outcomes that are not in the event. The event is 'precipitation occurs' (70% probability). The complement is 'precipitation does not occur'. The probability is $100% - 70% = 30%$ . 'No precipitation' could mean sunny, cloudy, or windy, so specifying 'sunny' or 'cloudy' is too narrow and incorrect.
The odds in favor of a team winning a championship are 3 to 7. What is the probability that the team does not win the championship?
Answer: $\frac{7}{10}$ Odds of 3 to 7 mean there are 3 favorable outcomes and 7 unfavorable outcomes, for a total of $3+7=10$ possible outcomes. The probability of the team not winning corresponds to the unfavorable outcomes. Therefore, the probability is $\frac{7}{10}$ .
A survey shows that 65% of students like pizza, 40% like burgers, and 25% like both pizza and burgers. If a student is selected at random and does NOT like pizza, what is the probability that this student likes burgers?
Answer: $\frac{3}{7}$ This is a conditional probability question involving set theory and survey data. When you see problems with overlapping groups (students who like both items), you need to carefully track what happens to each subset when conditions change. First, let's organize the given information. Out of all students: 65% like pizza, 40% like burgers, and 25% like both. This means 35% don't like pizza at all. Among those who don't like pizza, some will like burgers and some won't. To find how many students don't like pizza but do like burgers, subtract the overlap from the total burger lovers: 40% - 25% = 15% of all students like burgers but not pizza. Since 35% don't like pizza total, the probability that a non-pizza lover likes burgers is $\frac{15\%}{35\%} = \frac{15}{35} = \frac{3}{7}$ . Looking at the wrong answers: Choice A ( $\frac{4}{7}$ ) incorrectly uses the complement of our answer, perhaps from confusing conditional probability direction. Choice B ( $\frac{2}{5}$ ) appears to come from using the wrong denominator—possibly the total percentage who like burgers (40%) instead of those who don't like pizza. Choice C ( $\frac{15}{35}$ ) shows the unreduced fraction, which equals our answer but isn't simplified. For conditional probability problems involving surveys, always identify your new sample space first (in this case, the 35% who don't like pizza), then find what portion of that group meets your condition (the 15% who like burgers). Reduce fractions to match the answer format.
In a class of 25 students, 15 are taking biology and 12 are taking chemistry. If 8 students are taking both, what is the probability that a randomly selected student is taking neither biology nor chemistry?
Answer: $\frac{6}{25}$ First, find the number of students taking at least one of the subjects using the inclusion-exclusion principle: $N(\text{Biology} \cup \text{Chemistry}) = N(\text{Biology}) + N(\text{Chemistry}) - N(\text{Both}) = 15 + 12 - 8 = 19$ . This means 19 students are taking at least one of the courses. The number of students taking neither is the total number of students minus this amount: $25 - 19 = 6$ . The probability is therefore $\frac{6}{25}$ .
Two standard six-sided dice are rolled. What is the probability that the sum of the numbers rolled is not equal to 5?
Answer: $\frac{8}{9}$ There are $6 \times 6 = 36$ possible outcomes. The combinations that sum to 5 are (1, 4), (2, 3), (3, 2), and (4, 1). There are 4 such outcomes. The probability of the sum being 5 is $\frac{4}{36} = \frac{1}{9}$ . The probability of the sum not being 5 is the complement: $1 - \frac{1}{9} = \frac{8}{9}$ .
The probability that a certain machine part is defective is 0.03. What is the probability that a randomly selected part is not defective?
Answer: 0.97 The event that a part is not defective is the complement of the event that it is defective. If $P(\text{defective}) = 0.03$ , then $P(\text{not defective}) = 1 - P(\text{defective}) = 1 - 0.03 = 0.97$ .
In a game, the probability of winning on any single attempt is 0.3. A player continues until they either win or complete 3 unsuccessful attempts. What is the probability that the player does NOT win the game?
Answer: 0.343 When you encounter probability questions involving repeated attempts with stopping conditions, you need to identify all the ways the described outcome can occur and calculate their combined probability. The player does NOT win the game if they fail to win in their first attempt, fail again in their second attempt, and fail once more in their third attempt. Since the probability of winning is 0.3, the probability of losing on any single attempt is $1 - 0.3 = 0.7$ . For three consecutive losses, you multiply the individual probabilities: $0.7 \times 0.7 \times 0.7 = 0.7^3 = 0.343$ . Looking at the wrong answers: Choice A (0.700) represents the probability of losing just one attempt, not three consecutive attempts. Choice B (0.657) might come from incorrectly trying to add probabilities rather than multiply them, or from a flawed calculation involving the complement. Choice D (0.027) represents the probability of winning three times in a row ( $0.3^3$ ), which isn't relevant to this question. The correct answer is C (0.343). Remember that when dealing with independent events that must all occur (like losing three attempts in a row), you multiply their individual probabilities. Also, pay careful attention to stopping conditions in probability problems—the player stops after either winning once or losing three times, so "not winning the game" specifically means losing all three allowed attempts.
The probability that a company's stock price will increase is 0.45. The probability that it will decrease is 0.35. What is the probability that the stock price does not increase?
Answer: 0.55 The event is 'the stock price increases', with probability 0.45. The complement event is 'the stock price does not increase'. The probability of the complement is $1 - P(\text{increase}) = 1 - 0.45 = 0.55$ . The information about the price decreasing is part of the complement but not all of it, as the price could also remain unchanged.
The probability that a customer orders a drink is 0.8. The probability that a customer orders food is 0.6. The probability a customer orders both is 0.5. What is the probability that a randomly selected customer orders neither a drink nor food?
Answer: 0.1 The probability that a customer orders a drink or food is $P(D \cup F) = P(D) + P(F) - P(D \cap F) = 0.8 + 0.6 - 0.5 = 0.9$ . The event 'orders neither' is the complement of 'orders a drink or food'. Therefore, the probability of ordering neither is $1 - P(D \cup F) = 1 - 0.9 = 0.1$ .

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