Understand Compound Event Probability - 7th Grade Math

Two fair six-sided dice are rolled. A student counts only one favorable outcome for sum 7: $(1,6)$ . Which fraction correctly represents $P(\text{sum}=7)$ using all ordered outcomes?
Answer: $\frac{6}{36}$ This question tests compound event probability as the fraction of the compound sample space: list all possible outcomes, count the favorable ones, and divide by the total. A compound event combines simple events like two dice rolls, space 36 ordered pairs, P=favorable/total like coin both heads 1/4. For example, coins {HH,HT,TH,TT}, both heads 1/4; dice sum=7 has 6 favorable, 6/36=1/6. The correct fraction is 6/36 as there are six ordered pairs summing to 7, not just one. The student's error is counting only one outcome like (1,6), missing others like (6,1). To calculate, (1) identify two rolls, (2) space 36, (3) sum=7, (4) count 6, (5) P=6/36. Order matters, list systematically to avoid incomplete favorable counts.
Two fair coins are flipped. How many outcomes are in the compound sample space when order matters (first coin, then second coin)?
Answer: 4 This question tests compound event probability as the fraction of the compound sample space by listing outcomes, counting favorable ones, and dividing by the total. A compound event combines simple events like flipping two coins, with the compound space including all combinations such as {HH, HT, TH, TT}=4 for two coins or 36 for two dice. The event probability is found by identifying favorable outcomes and calculating P = favorable/total, like both heads {HH} with P=1/4. For example, with two coins {HH, HT, TH, TT}, 'both heads' has {HH}, so P=1/4; for sum=7 on dice, there are 6 favorable outcomes, P=6/36=1/6. The correct answer is C, 4, because the sample space has 4 outcomes when order matters: HH, HT, TH, TT. A common error is treating order as not mattering, leading to only 3 outcomes like both heads, one each, both tails. To find this, (1) identify the compound process of two coin flips, (2) list or count the space as 4, (3) identify favorable (not needed here), (4) count total, (5) no P but total is key; systematic listing helps, and order matters. Mistakes include incomplete space like missing TH or HT, or ignoring order.
Elena flips three fair coins simultaneously. What is the probability that she gets exactly two heads, given that she gets at least one head?
Answer: $\frac{3}{7}$ This is conditional probability. P(exactly 2 heads | at least 1 head) = P(exactly 2 heads AND at least 1 head) ÷ P(at least 1 head). Since exactly 2 heads automatically satisfies at least 1 head, the numerator is just P(exactly 2 heads) = 3/8. P(at least 1 head) = 1 - P(no heads) = 1 - 1/8 = 7/8. Therefore, P(exactly 2 heads | at least 1 head) = (3/8) ÷ (7/8) = 3/7. Choice A ignores the condition. Choice C incorrectly assumes equal likelihood among favorable outcomes. Choice D miscounts favorable outcomes.
Two fair six-sided dice are rolled. What is the probability that the first die shows 1 and the sum is 7?
Answer: $\frac{1}{36}$ Since the first die must show a 1 and the sum of both dice must be 7, the second die has to show a 6, so only the single outcome, 1 and 6, out of all 36 possible dice combinations satisfies both conditions, giving a probability of 1/36, matching choice D. Choice B, 6/36 or 1/6, is the probability of getting a sum of 7 in general, from any of the 6 combinations that sum to 7, without the added condition that the first die specifically shows a 1. Choice C, 1/12, does not correspond to a correct count of favorable outcomes over the 36 total outcomes. Choice A, 1/11, would come from mistakenly treating the possible sums 2 through 12 as 11 equally likely outcomes, which is not how dice probabilities actually work. Listing out the 36 equally likely ordered pairs is the most reliable way to confirm that only one outcome meets both conditions in this problem.
Two fair six-sided dice are rolled. What is the probability that the sum is 7, written as a simplified fraction?
Answer: $\frac{1}{6}$ This question tests compound event probability as the fraction of the compound sample space: list all possible outcomes, count the favorable ones, and divide by the total. A compound event combines simple events like two dice, space 36, P=favorable/total like coins both heads 1/4. For example, coins {HH,HT,TH,TT}, both heads 1/4; dice sum=7 has (1,6) to (6,1), 6 favorable, P=6/36=1/6. The correct simplified fraction is 1/6 since 6/36 reduces to 1/6. A common error is miscounting favorable to 7, giving 7/36. To calculate, (1) identify rolls, (2) space 36, (3) sum=7, (4) count 6, (5) P=6/36=1/6. Order matters, list pairs systematically to avoid over or undercounting.
A fair coin is flipped and then a fair six-sided die is rolled. How many favorable outcomes are there for the event "heads and even"?
Answer: 3 There are 3 even numbers on the die (2, 4, 6), and only one way to get heads, so "heads and even" has 3 favorable outcomes: H2, H4, H6, matching choice A. Choice B (12) is the total number of outcomes in the full sample space (2 coin faces x 6 die faces), not the number of favorable outcomes. Choice C (6) mistakenly counts all six die faces paired with heads, both even and odd. Choice D (2) undercounts, missing one of the even outcomes.
A game involves rolling a standard die twice. Players win if they get a sum of 7 or if both rolls show even numbers. What is the probability of winning this game?
Answer: $\frac{5}{12}$ When you see probability questions involving "or" conditions, you need to be careful about overlapping outcomes. This game has two winning conditions: getting a sum of 7 OR getting both even numbers. Let's find each probability separately, then account for overlap. With two dice rolls, there are 36 total possible outcomes. For a sum of 7, the winning combinations are: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) — that's 6 outcomes, so P(sum of 7) = $\frac{6}{36} = \frac{1}{6}$ . For both rolls even, you need even numbers (2, 4, 6) on both dice. That's 3 × 3 = 9 outcomes, so P(both even) = $\frac{9}{36} = \frac{1}{4}$ . Here's the key: these events can't happen simultaneously. A sum of 7 requires one odd and one even number, while "both even" requires... both even. Since there's no overlap, you simply add the probabilities: $\frac{1}{6} + \frac{1}{4} = \frac{2}{12} + \frac{3}{12} = \frac{5}{12}$ . Choice A ( $\frac{1}{3}$ ) likely comes from incorrectly counting favorable outcomes. Choice B ( $\frac{4}{9}$ ) might result from using 18 as the denominator instead of 36. Choice C ( $\frac{7}{18}$ ) could come from miscalculating the overlap or the individual probabilities. Remember: when dealing with "or" in probability, add the individual probabilities but subtract any overlap. Always check whether the events can occur simultaneously — it determines whether you need that subtraction step.
A student flips a fair coin and then rolls a fair six-sided die. The compound sample space has $2\times 6=12$ equally likely outcomes (like $H1, H2, \dots, T6$ ). What is the probability of getting heads and an even number?
Answer: $\frac{3}{12}=\frac{1}{4}$ This question tests compound event probability as the fraction of the compound sample space by listing outcomes, counting favorable ones, and dividing by the total. A compound event combines simple events like flipping a coin and rolling a die, with the compound space including all combinations such as {HH, HT, TH, TT}=4 for two coins or 36 for two dice. The event probability is found by identifying favorable outcomes and calculating P = favorable/total, like both heads {HH} with P=1/4. For example, with two coins {HH, HT, TH, TT}, 'both heads' has {HH}, so P=1/4; for sum=7 on dice, there are 6 favorable outcomes, P=6/36=1/6. The correct answer is A, $\frac{3}{12}=\frac{1}{4}$ , because there are 3 favorable outcomes (H2, H4, H6) out of 12 total. A common error is using only the die's space (6) instead of the compound 12, leading to wrong fractions like 3/6. To find this, (1) identify the compound process of coin and die, (2) list or count the space as 12, (3) identify favorable as heads and even, (4) count 3, (5) calculate P=3/12; systematic listing helps, and order matters as coin first, die second. Mistakes include incomplete space, miscounted favorable like 2 or 6, or using simple probability.
Two fair six-sided dice are rolled. How many outcomes give a sum of 7?
Answer: $6$ This question tests compound event probability as the fraction of the compound sample space by listing outcomes, counting favorable ones, and dividing by the total. A compound event combines simple events like rolling two dice, with the compound space including all combinations for a total of 36 outcomes, and the probability is calculated by identifying favorable outcomes and dividing by the total, for example, both heads on two coins {HH} out of {HH, HT, TH, TT} gives $P=1/4$ . For example, with two coins {HH, HT, TH, TT}, the event 'both heads' has {HH} as favorable, so $P=1/4$ ; similarly, for sum=7 on two dice, there are 6 favorable outcomes out of 36, so $P=6/36=1/6$ . The correct answer is B, 6, because the pairs $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$ sum to 7. A common error is miscounting by ignoring order, like thinking only 3 unique pairs. To find such probabilities, (1) identify the compound process, (2) list or count the sample space, (3) identify favorable outcomes, (4) count them, and (5) calculate $P$ as favorable over total. Use systematic listing to ensure all outcomes are included, remembering that order matters in distinguishing $(1,6)$ from $(6,1)$ , and avoid mistakes like miscounted favorable or order ignored.
A student flips two fair coins. The compound sample space is {HH, HT, TH, TT}. What is the probability of getting two heads?
Answer: 1/4 Flipping two coins gives four equally likely outcomes: HH, HT, TH, and TT. Only one of these outcomes, HH, shows two heads, so the probability is 1 out of 4, or 1/4, matching Choice A. Choice B and Choice C both give 1/2, which would be the probability of getting exactly one head, not two heads. Choice D, 1/3, does not match any correct count from the sample space of 4 outcomes.
Carlos rolls two standard six-sided dice. Given that the sum of the dice is greater than 8, what is the probability that both dice show the same number? Give your answer in simplest form.
Answer: $\frac{1}{5}$ This is conditional probability. First, find outcomes where sum > 8: (3,6), (4,5), (4,6), (5,4), (5,5), (5,6), (6,3), (6,4), (6,5), (6,6) = 10 outcomes. Among these, the outcomes where both dice match are: (5,5), (6,6) = 2 outcomes. P(both same | sum > 8) = 2/10 = 1/5. Choice B is the unreduced fraction. Choice C ignores the condition and uses 6/36. Choice D incorrectly counts outcomes where sum > 8 and includes non-matching pairs.
A student flips two fair coins. Which list correctly shows the entire compound sample space?
Answer: {HH, HT, TH, TT} When you flip two coins, each coin can land on heads or tails, so there are 2 x 2 = 4 possible outcomes total. Listing every outcome gives {HH, HT, TH, TT}, which matches Choice A. Choice B is missing HT and TH, so it leaves out two possible outcomes. Choice C only shows the outcomes for one coin, not two. Choice D is missing TH, so it does not show every way the coins could land.
Use the table shown to answer the question. In a survey of 120 students about their favorite subjects, what is the probability that a randomly selected student who likes Math also likes Science?
Answer: $\frac{25}{45}$ This asks for P(likes Science | likes Math). From the table, 45 students like Math total, and 25 of those also like Science. So P(likes Science | likes Math) = 25/45. Choice A ignores the condition and uses the total population. Choice C uses the wrong numerator (students who like Science but not Math). Choice D uses the wrong denominator (total Science lovers).
A student flips a fair coin and rolls a fair six-sided die. How many outcomes are favorable for the event "heads and even"?
Answer: 3 This question tests compound event probability as the fraction of the compound sample space: list all possible outcomes, count the favorable ones, and divide by the total. A compound event combines simple events like coin flip and die roll, with compound space of 12 outcomes, probability as favorable over total, like both heads on two coins P=1/4. For example, two coins {HH, HT, TH, TT}, both heads {HH}, P=1/4; sum=7 on dice, 6 favorable, P=6/36=1/6. The correct number is 3 because favorable are H with even die: H2, H4, H6. A common error is counting all evens including tails, leading to 6. To find the count, (1) identify compound process, (2) list space of 12, (3) identify heads and even, (4) count 3 (evens:2,4,6 with H), (5) for P=3/12. Systematic listing by coin then die ensures accuracy; mistakes include miscounting evens or including tails.
Two fair coins are flipped. What is the probability of getting exactly one head?
Answer: $\frac{1}{2}$ This question tests compound event probability as the fraction of the compound sample space by listing outcomes, counting favorable ones, and dividing by the total. A compound event combines simple events like flipping two coins, with the compound space including all combinations such as {HH, HT, TH, TT}=4 for two coins or 36 for two dice. The event probability is found by identifying favorable outcomes and calculating P = favorable/total, like both heads {HH} with P=1/4. For example, with two coins {HH, HT, TH, TT}, 'both heads' has {HH}, so P=1/4; for sum=7 on dice, there are 6 favorable outcomes, P=6/36=1/6. The correct answer is B, $\frac{1}{2}$ , because there are 2 favorable (HT, TH) out of 4, so 2/4=1/2. A common error is counting only one way for one head, leading to 1/4 or 3/4. To find this, (1) identify the compound process of two coins, (2) list or count the space as 4, (3) identify favorable as exactly one head, (4) count 2, (5) calculate P=2/4=1/2; systematic listing helps, and order matters. Mistakes include incomplete space or miscounted favorable like 3.
Two fair coins are flipped. How many outcomes are in the compound sample space when order matters (first coin, second coin)?
Answer: 4 This question tests compound event probability as the fraction of the compound sample space: list all possible outcomes, count the favorable ones, and divide by the total. A compound event combines simple events like flipping two coins, with the compound space being all combinations {HH, HT, TH, TT} totaling 4 when order matters, and probability calculated as favorable over total, like both heads {HH} giving $P=1/4$ . For example, with two coins {HH, HT, TH, TT}, 'both heads' has 1 favorable, $P=1/4$ ; for sum=7 on dice, 6 favorable out of 36, $P=6/36=1/6$ . The correct answer is 4 because ordered outcomes distinguish first and second coin, giving HH, HT, TH, TT. A common error is thinking it's 3 by treating HT and TH as the same, ignoring order. To find the sample space size, (1) identify the compound process of two flips, (2) list or count space as $2x2=4$ , (3) no event here but for probability would identify favorable, (4) count them, (5) calculate P if needed. Remember order matters in compound spaces; mistakes include incomplete listing or assuming unordered outcomes.
Two fair six-sided dice are rolled. There are 36 outcomes in the compound sample space. What is the probability that the sum is 7?
Answer: 1/6 When two dice are rolled, there are 36 total equally likely outcomes. The pairs that sum to 7 are (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1), which is 6 favorable outcomes. Dividing 6 by 36 gives 6/36, which simplifies to 1/6, matching Choice B. Choice A, 1/36, is the probability of rolling one specific pair, not all six ways to get a sum of 7. Choice C and Choice D do not match the correct count of favorable outcomes over the total sample space.
Two fair six-sided dice are rolled. How many outcomes are in the compound sample space of ordered pairs $(a,b)$ ?
Answer: 36 This question tests compound event probability as the fraction of the compound sample space by listing outcomes, counting favorable ones, and dividing by the total. A compound event combines simple events like rolling two dice, with the compound space including all combinations such as {HH, HT, TH, TT}=4 for two coins or 36 for two dice. The event probability is found by identifying favorable outcomes and calculating P = favorable/total, like both heads {HH} with P=1/4. For example, with two coins {HH, HT, TH, TT}, 'both heads' has {HH}, so P=1/4; for sum=7 on dice, there are 6 favorable outcomes, P=6/36=1/6. The correct answer is C, 36, because there are 6 options for each die, so 6x6=36 ordered pairs. A common error is adding instead of multiplying, like 6+6=12, or doubling to 72. To find this, (1) identify the compound process of two dice, (2) list or count the space as 36, (3) identify favorable (not here), (4) count total, (5) no P; systematic listing helps, and order matters for pairs. Mistakes include incomplete space like 18 for unordered, or miscounting as 12.
A weather app shows that the probability of rain on Saturday is 0.3 and the probability of rain on Sunday is 0.4. Assuming these events are independent, what is the probability that it rains on exactly one of these two days?
Answer: $0.46$ P(exactly one day rains) = P(rain Sat, no rain Sun) + P(no rain Sat, rain Sun) = (0.3)(0.6) + (0.7)(0.4) = 0.18 + 0.28 = 0.46. Choice A represents P(rain both days) = 0.3 × 0.4. Choice C represents P(at least one day rains) = 1 - P(no rain either day) = 1 - 0.7 × 0.6 = 0.58. Choice D incorrectly adds the individual probabilities without considering independence.
A student flips two fair coins. The compound sample space is $\{HH, HT, TH, TT\}$ . What is the probability of getting two heads?
Answer: $\frac{1}{4}$ This question tests compound event probability as the fraction of the compound sample space: list all possible outcomes, count the favorable ones, and divide by the total. A compound event combines simple events like flipping two coins, with the compound space being all combinations such as {HH, HT, TH, TT} totaling 4 outcomes, and the probability is the number of favorable outcomes divided by total, like both heads {HH} giving P=1/4. For example, with two coins {HH, HT, TH, TT}, the event 'both heads' has {HH} as favorable, so P=1/4; similarly, for sum=7 on two dice, there are 6 favorable out of 36, P=6/36=1/6. The correct answer is 1/4 because there is only one favorable outcome, HH, out of four equally likely outcomes. A common error is thinking it's 1/2 by mistakenly using single coin probability without considering the compound space. To find this probability, (1) identify the compound process of two coin flips, (2) list or count the space as 4 outcomes, (3) identify favorable as two heads, (4) count 1 favorable, (5) calculate P=1/4. Use systematic listing to ensure all outcomes are included, remembering order matters in distinguishing HT and TH; mistakes include incomplete spaces or ignoring order.

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