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Question 1 / 20
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1
What is the formula for calculating the mean of a binomial distribution?
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Solution: Step 1: Recall the standard formulas for the parameters of a binomial distribution. Step 2: For a binomial distribution with 'n' trials and probability of success 'p' in each trial, the mean (expected value) is given by the product of 'n' and 'p'. Step 3: The formula for the mean of a binomial distribution is μ = np.
2
A box contains 8 red, 7 blue, and 6 green balls. If one ball is chosen at random, what is the probability that this ball is neither blue nor green?
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Solution: Step 1: Calculate the total number of balls in the box. Total balls = 8 (red) + 7 (blue) + 6 (green) = 21 balls. Step 2: Identify the condition 'neither blue nor green'. If a ball is neither blue nor green, it must be red. Step 3: Determine the number of favorable outcomes (Event E: drawing a red ball). Number of red balls = 8. So, n(E) = 8. Step 4: Calculate the probability of event E. P(E) = n(E) / Total balls = 8 / 21.
3
When two coins are simultaneously tossed, what is the probability of obtaining at least one tail?
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Solution: Step 1: List all possible outcomes when two coins are tossed simultaneously. The sample space (S) = {HH, HT, TH, TT}. Step 2: Count the total number of outcomes. n(S) = 4. Step 3: Identify the outcomes where there is at least one tail. Favorable outcomes (E) = {HT, TH, TT}. Step 4: Count the number of favorable outcomes. n(E) = 3. Step 5: Calculate the probability. P(E) = n(E) / n(S) = 3/4.
4
In a test taken by 600 students, 35% failed Physics and 45% failed Chemistry. If 40% of those who passed Chemistry also passed Physics, how many students failed both subjects?
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Solution: Step 1: Total students = 600. Step 2: Percentage of students who passed Physics = 100% - 35% = 65%. Step 3: Percentage of students who passed Chemistry = 100% - 45% = 55%. Step 4: Number of students who passed Chemistry = 55% of 600 = 0.55 × 600 = 330. Step 5: Number of students who passed both Physics and Chemistry = 40% of (those who passed Chemistry) = 0.40 × 330 = 132. Step 6: Percentage of students who passed in both subjects = (132 / 600) × 100% = 22%. Step 7: Using the Principle of Inclusion-Exclusion for percentages (P(A U B) = P(A) + P(B) - P(A ∩ B)), calculate the percentage of students who passed at least one subject: * Passed at least one = 65% (Physics) + 55% (Chemistry) - 22% (Both) = 98%. Step 8: The percentage of students who failed in both subjects is 100% - (Percentage who passed at least one subject) = 100% - 98% = 2%. Step 9: Calculate the number of students who failed in both subjects: 2% of 600 = 0.02 × 600 = 12 students.
5
From a standard deck of 52 cards, one card is drawn. What is the probability that this card is either a diamond or a king?
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Solution: Step 1: Determine the total number of cards in a standard deck: n(S) = 52. Step 2: Determine the number of diamond cards: There are 13 diamonds. Step 3: Determine the number of king cards: There are 4 kings. Step 4: Identify the overlap: One card is both a diamond and a king (King of Diamonds). Step 5: Calculate the number of cards that are a diamond OR a king using the principle of inclusion-exclusion: n(Diamond ∪ King) = n(Diamond) + n(King) - n(Diamond ∩ King) = 13 + 4 - 1 = 16. Step 6: Calculate the probability P(Diamond or King) = n(Diamond ∪ King) / n(S) = 16 / 52. Step 7: Simplify the probability: 16 / 52 = 4 / 13.
6
When two dice are thrown, what is the probability that their total score is a prime number?
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Solution: Step 1: Determine the total number of possible outcomes when two dice are thrown. n(S) = 6 * 6 = 36. Step 2: Identify the prime numbers that can be formed by the sum of two dice (minimum 2, maximum 12). Prime sums: {2, 3, 5, 7, 11}. Step 3: List the outcomes for each prime sum: Sum = 2: {(1, 1)} (1 outcome) Sum = 3: {(1, 2), (2, 1)} (2 outcomes) Sum = 5: {(1, 4), (2, 3), (3, 2), (4, 1)} (4 outcomes) Sum = 7: {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} (6 outcomes) Sum = 11: {(5, 6), (6, 5)} (2 outcomes) Step 4: Calculate the total number of favorable outcomes (sum is a prime number). n(E) = 1 + 2 + 4 + 6 + 2 = 15. Step 5: Calculate the probability. P(E) = n(E) / n(S) = 15 / 36. Step 6: Simplify the fraction. P(E) = 5 / 12.
7
A husband and wife are interviewed for two open positions in the same role. The husband has a selection probability of 1/7, and the wife has a selection probability of 1/5. What is the probability that exactly one of them is chosen?
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Solution: Step 1: Define the probabilities of selection for the husband and wife. Let P(H) = Probability of husband's selection = 1/7. Let P(W) = Probability of wife's selection = 1/5. Step 2: Calculate the probabilities of non-selection for both. P(H') = Probability of husband not being selected = 1 - 1/7 = 6/7. P(W') = Probability of wife not being selected = 1 - 1/5 = 4/5. Step 3: Identify the two scenarios where only one of them is selected: Scenario 1: Husband is selected AND Wife is not selected. Scenario 2: Wife is selected AND Husband is not selected. Step 4: Calculate the probability for Scenario 1 (assuming independence). P(H and W') = P(H) * P(W') = (1/7) * (4/5) = 4/35. Step 5: Calculate the probability for Scenario 2 (assuming independence). P(W and H') = P(W) * P(H') = (1/5) * (6/7) = 6/35. Step 6: Since these two scenarios are mutually exclusive, add their probabilities to find the total probability that only one is selected. P(only one selected) = P(H and W') + P(W and H') = 4/35 + 6/35 = 10/35 = 2/7.
8
If a single die is thrown twice, what is the probability that the sum of the digits (numbers) appearing on the two rolls is 10?
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Solution: Step 1: Calculate the total number of possible outcomes when throwing a die twice. Each throw has 6 outcomes, so total outcomes = 6 * 6 = 36. Step 2: Identify all combinations of two dice rolls that result in a sum of 10. These combinations are: (4, 6), (5, 5), (6, 4). Step 3: Count the total number of favorable outcomes: 3. Step 4: Calculate the probability: (Favorable Outcomes) / (Total Outcomes) = 3 / 36. Step 5: Simplify the fraction: 3/36 = 1/12.
9
A single card is randomly selected from a deck of 52 cards. What is the probability that the chosen card is a face card?
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Solution: Step 1: Identify the total number of possible outcomes (sample space). Total number of cards = 52. So, n(S) = 52. Step 2: Identify the number of face cards in a standard deck. Face cards include Jacks, Queens, and Kings. There are 4 of each. Total number of face cards = 4 + 4 + 4 = 12. So, n(E) = 12. Step 3: Calculate the probability. P(E) = n(E) / n(S) = 12 / 52. Step 4: Simplify the fraction. P(E) = 3 / 13.
10
Two unbiased coins are tossed. What is the probability of obtaining at most one tail?
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Solution: Step 1: List all possible outcomes when two coins are tossed simultaneously. The sample space (S) = {HH, HT, TH, TT}. Step 2: Count the total number of outcomes. n(S) = 4. Step 3: Identify the outcomes where there is 'at most one tail' (meaning zero tails or one tail). Favorable outcomes (E) = {HH, HT, TH}. Step 4: Count the number of favorable outcomes. n(E) = 3. Step 5: Calculate the probability. P(E) = n(E) / n(S) = 3/4.
11
A box contains 6 black, 4 red, 2 white, and 3 blue shirts. If 3 shirts are randomly selected from the box, what is the probability of choosing 2 red shirts and 1 blue shirt?
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Solution: Step 1: Calculate the total number of shirts. Total shirts = 6 (black) + 4 (red) + 2 (white) + 3 (blue) = 15 shirts. Step 2: Calculate the total number of ways to choose 3 shirts from 15. This is C(15, 3) = (15 * 14 * 13) / (3 * 2 * 1) = 455 ways. Step 3: Calculate the number of ways to choose 2 red shirts from 4 red shirts. C(4, 2) = (4 * 3) / (2 * 1) = 6 ways. Step 4: Calculate the number of ways to choose 1 blue shirt from 3 blue shirts. C(3, 1) = 3 ways. Step 5: Calculate the number of favorable outcomes (choosing 2 red and 1 blue). Multiply the ways from Step 3 and Step 4: 6 * 3 = 18 ways. Step 6: Calculate the probability. P(2 red, 1 blue) = (Favorable outcomes) / (Total outcomes) = 18/455.
12
For the special rule of multiplication in probability to apply, what condition must the events satisfy?
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Solution: Step 1: Recall the rules of probability multiplication. Step 2: The general multiplication rule P(A and B) = P(A) × P(B|A) applies to all events. Step 3: The *special* multiplication rule P(A and B) = P(A) × P(B) is a simplified form that applies specifically when events A and B are independent. Step 4: Therefore, for the special rule of multiplication of probability, the events must be 'Independent'.
13
A card is drawn randomly from a standard deck of 52 cards. What is the probability that the drawn card is neither a spade nor a Jack?
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Solution: Step 1: Determine the total number of cards in a standard deck: n(S) = 52. Step 2: Calculate the probability of the complementary event: "getting a spade OR a Jack." Step 3: Count the number of spades: 13. Step 4: Count the number of Jacks: 4. Step 5: Identify the overlap: There is 1 card that is both a spade and a Jack (Jack of Spades). Step 6: Calculate the number of cards that are a spade OR a Jack: n(Spade ∪ Jack) = n(Spade) + n(Jack) - n(Spade ∩ Jack) = 13 + 4 - 1 = 16. Step 7: Calculate the probability of getting a spade or a Jack: P(Spade ∪ Jack) = 16 / 52 = 4 / 13. Step 8: Calculate the probability of getting neither a spade nor a Jack (complementary event): P(Neither Spade nor Jack) = 1 - P(Spade ∪ Jack) = 1 - 4/13 = (13 - 4) / 13 = 9 / 13.
14
Determine the probability of a non-leap year containing 53 Thursdays.
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Solution: Step 1: Identify the number of days in a non-leap year. A non-leap year has 365 days. Step 2: Convert days to weeks and remaining days. 365 days = 52 weeks and 1 extra day (365 = 52 * 7 + 1). Step 3: Recognize that 52 weeks guarantee 52 Thursdays. The probability of having 53 Thursdays depends on this one extra day. Step 4: List all possible days for the extra day. The extra day can be any of Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday (7 possibilities). Step 5: Determine the favorable outcome. For 53 Thursdays, the extra day must be a Thursday (1 favorable outcome). Step 6: Calculate the probability. P(53 Thursdays) = (Favorable outcomes) / (Total possible outcomes) = 1/7.
15
When two dice are thrown at the same time, what is the probability of getting a total sum of 7?
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Solution: Step 1: Determine the total number of outcomes when two dice are thrown simultaneously (Sample Space, n(S)). n(S) = 6 (outcomes for first die) × 6 (outcomes for second die) = 36. Step 2: Identify the favorable outcomes where the sum of the face numbers is 7. E = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)} Step 3: Count the number of favorable outcomes, n(E). n(E) = 6. Step 4: Calculate the probability of event E. P(E) = n(E) / n(S) = 6 / 36. Step 5: Simplify the probability. P(E) = 1 / 6.
16
If two dice are rolled at the same time, what is the probability that the product of the numbers shown on their faces is an even number?
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Solution: Step 1: Determine the total number of possible outcomes when two dice are thrown. n(S) = 6 * 6 = 36. Step 2: Understand the condition for the product of two numbers to be even. The product of two numbers is even if at least one of the numbers is even. The product is odd ONLY if both numbers are odd. Step 3: Use the complementary probability approach. The complement of "product is even" is "product is odd". Step 4: List the outcomes where both numbers are odd. Odd numbers on a die: {1, 3, 5}. Outcomes where both are odd: {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)}. Number of outcomes where the product is odd, n(E') = 9. Step 5: Calculate the probability of the complementary event (product is odd). P(E') = n(E') / n(S) = 9 / 36 = 1 / 4. Step 6: Calculate the required probability (product is even). P(product is even) = 1 - P(product is odd) = 1 - (1/4) = 3 / 4.
17
A box contains nine light bulbs, with 4 of them being defective. If four bulbs are randomly selected, what is the probability that exactly three of these bulbs are good?
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Solution: Step 1: Determine the total number of bulbs: 9. Step 2: Identify the number of defective bulbs (4) and good bulbs (9 - 4 = 5). Step 3: Calculate the total number of ways to choose 4 bulbs from 9: 9 C 4 = (9 * 8 * 7 * 6) / (4 * 3 * 2 * 1) = 126 ways. Step 4: For 'exactly three bulbs are good', we must select 3 good bulbs AND 1 defective bulb. Step 5: Calculate the number of ways to choose 3 good bulbs from 5: 5 C 3 = (5 * 4 * 3) / (3 * 2 * 1) = 10 ways. Step 6: Calculate the number of ways to choose 1 defective bulb from 4: 4 C 1 = 4 ways. Step 7: Find the total number of favorable ways (3 good and 1 defective) = (ways to choose good) * (ways to choose defective) = 10 * 4 = 40 ways. Step 8: Calculate the probability: (Favorable Outcomes) / (Total Outcomes) = 40 / 126. Step 9: Simplify the fraction: 40/126 = 20/63.
18
From a group of 3 men and 2 women, a committee of 3 members is to be selected. What is the probability that this committee includes at least one woman?
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Solution: Step 1: Calculate the total number of persons: 3 men + 2 women = 5 persons. Step 2: Calculate the total number of ways to select a 3-member committee from 5 persons: n(S) = ⁵C₃ = (5 × 4 × 3) / (3 × 2 × 1) = 10. Step 3: Determine the number of ways to select a committee with no women (i.e., all 3 members are men): ³C₃ = 1. Step 4: Calculate the number of favorable outcomes (at least 1 woman) using the complementary event: n(E) = Total ways - Ways with no women = 10 - 1 = 9. Step 5: Calculate the probability P(E) = n(E) / n(S) = 9 / 10.
19
A single die is cast twice. Given that the sum of the numbers appearing on the two rolls is 10, what is the probability that the number 5 has appeared at least once?
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Solution: Step 1: First, identify all possible outcomes when a die is cast twice, such that their sum is 10. These outcomes are: (4, 6), (5, 5), (6, 4). Step 2: The total number of outcomes in this reduced sample space is 3. Step 3: From these outcomes, identify which ones include the number 5 appearing at least once. Only the outcome (5, 5) satisfies this condition. Step 4: The number of favorable outcomes (where 5 appears at least once among the sums of 10) is 1. Step 5: Calculate the conditional probability: (Favorable Outcomes) / (Total Outcomes in Reduced Sample Space) = 1 / 3.
20
The index numbers for five items are 121, 123, 125, 126, 128, with respective weights of 5, 11, 10, 8, 6. What is the weighted average index number?
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Solution: Step 1: Calculate the summation of the product of Index and Weights = 121*5 + 123*11 + 125*10 + 126*8 + 128*6 = 605 + 1353 + 1250 + 1008 + 768 = 4984 Step 2: Calculate the summation of Weights = 5 + 11 + 10 + 8 + 6 = 40 Step 3: Calculate the weighted average = 4984 / 40 = 124.6
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