📘 Quiz

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Question 1 / 20
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1
There are two pots. Pot 1 contains 5 red and 3 green marbles. Pot 2 contains 4 red and 2 green marbles. If one pot is chosen randomly, and then one marble is drawn from it, what is the probability of drawing a red marble?
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Solution: Step 1: Determine the probability of selecting each pot. * P(Select Pot 1) = 1/2. * P(Select Pot 2) = 1/2. Step 2: Calculate the conditional probability of drawing a red marble from Pot 1. * Pot 1 has 5 red + 3 green = 8 marbles. * P(Red | Pot 1) = 5/8. Step 3: Calculate the conditional probability of drawing a red marble from Pot 2. * Pot 2 has 4 red + 2 green = 6 marbles. * P(Red | Pot 2) = 4/6 = 2/3. Step 4: Apply the Law of Total Probability: P(Red) = P(Red | Pot 1) * P(Select Pot 1) + P(Red | Pot 2) * P(Select Pot 2). Step 5: Substitute the values: P(Red) = (5/8) * (1/2) + (2/3) * (1/2). Step 6: Simplify: P(Red) = 5/16 + 2/6 = 5/16 + 1/3. Step 7: Find a common denominator (48) and sum the fractions: * P(Red) = (5 * 3) / (16 * 3) + (1 * 16) / (3 * 16) * P(Red) = 15/48 + 16/48 = 31/48.
2
A box contains 20 electrical bulbs, 4 of which are faulty. If two bulbs are chosen randomly from this box, what is the probability that at least one of the selected bulbs is defective?
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Solution: Step 1: Identify the total number of bulbs and the number of defective/non-defective bulbs. Total bulbs = 20. Defective bulbs = 4. Non-defective (good) bulbs = 20 - 4 = 16. Step 2: Calculate the total number of ways to choose 2 bulbs from 20 (sample space). n(S) = 20 C 2 = (20 * 19) / (2 * 1) = 190. Step 3: Use the complementary probability approach. The complement of "at least one is defective" is "none are defective". Step 4: Calculate the number of ways to choose 2 bulbs such that none are defective (i.e., both are good). n(E') = 16 C 2 = (16 * 15) / (2 * 1) = 120. Step 5: Calculate the probability of the complementary event (none are defective). P(E') = n(E') / n(S) = 120 / 190 = 12 / 19. Step 6: Calculate the required probability (at least one is defective). P(at least one defective) = 1 - P(none are defective) = 1 - (12/19) = 7 / 19.
3
A basket contains 5 apples and 4 oranges. Three fruits are randomly selected. What is the probability that at least 2 apples are picked?
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Solution: Step 1: Calculate the total number of fruits in the basket: 5 apples + 4 oranges = 9 fruits. Step 2: Determine the total number of ways to pick 3 fruits from 9: 9 C 3 = (9 * 8 * 7) / (3 * 2 * 1) = 84 ways. Step 3: 'At least 2 apples' means either exactly 2 apples OR exactly 3 apples are picked. Step 4: Case 1: Exactly 2 apples and 1 orange. * Ways to choose 2 apples from 5: 5 C 2 = (5 * 4) / (2 * 1) = 10 ways. * Ways to choose 1 orange from 4: 4 C 1 = 4 ways. * Number of ways for Case 1 = 10 * 4 = 40 ways. Step 5: Case 2: Exactly 3 apples and 0 oranges. * Ways to choose 3 apples from 5: 5 C 3 = (5 * 4 * 3) / (3 * 2 * 1) = 10 ways. * Ways to choose 0 oranges from 4: 4 C 0 = 1 way. * Number of ways for Case 2 = 10 * 1 = 10 ways. Step 6: Total favorable outcomes = (Ways for Case 1) + (Ways for Case 2) = 40 + 10 = 50 ways. Step 7: Calculate the probability: (Favorable Outcomes) / (Total Outcomes) = 50 / 84. Step 8: Simplify the fraction: 50/84 = 25/42.
4
A single die is rolled two times. What is the probability that the sum of the numbers shown on the two rolls is 9?
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Solution: Step 1: Determine the total number of possible outcomes when a die is rolled twice: 6 outcomes for the first roll * 6 outcomes for the second roll = 36 total outcomes. Step 2: Identify the pairs of outcomes where the sum of the two rolls is 9: * (3, 6) * (4, 5) * (5, 4) * (6, 3) Step 3: The number of favorable outcomes is 4. Step 4: Calculate the probability: (Favorable Outcomes) / (Total Outcomes) = 4 / 36. Step 5: Simplify the fraction: 4/36 = 1/9.
5
An urn holds 6 red, 4 blue, 2 green, and 3 yellow marbles. If four marbles are randomly selected, what is the probability that the selection consists of 1 green, 2 blue, and 1 red marble?
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Solution: Step 1: Calculate the total number of marbles in the urn. Total marbles = 6 (red) + 4 (blue) + 2 (green) + 3 (yellow) = 15 marbles. Step 2: Calculate the total number of ways to pick 4 marbles from 15 (sample space). n(S) = 15 C 4 = (15 * 14 * 13 * 12) / (4 * 3 * 2 * 1) = 1365. Step 3: Calculate the number of ways to pick 1 green marble from 2 green marbles. n(Green) = 2 C 1 = 2. Step 4: Calculate the number of ways to pick 2 blue marbles from 4 blue marbles. n(Blue) = 4 C 2 = (4 * 3) / (2 * 1) = 6. Step 5: Calculate the number of ways to pick 1 red marble from 6 red marbles. n(Red) = 6 C 1 = 6. Step 6: Calculate the total number of favorable outcomes (1 green, 2 blue, 1 red). Multiply the combinations: n(E) = n(Green) * n(Blue) * n(Red) = 2 * 6 * 6 = 72. Step 7: Calculate the probability. P(E) = n(E) / n(S) = 72 / 1365. Step 8: Simplify the fraction. P(E) = 24 / 455.
6
A box contains 6 black, 4 red, 2 white, and 3 blue shirts. What is the probability of randomly drawing 2 black shirts?
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Solution: Step 1: Calculate the total number of shirts in the box. Total shirts = 6 (black) + 4 (red) + 2 (white) + 3 (blue) = 15 shirts. Step 2: Calculate the probability of drawing the first black shirt. P(1st black) = 6/15. Step 3: After drawing one black shirt (without replacement), calculate the remaining shirts and black shirts. Remaining black shirts = 5. Remaining total shirts = 14. Step 4: Calculate the probability of drawing the second black shirt. P(2nd black | 1st black) = 5/14. Step 5: Multiply the probabilities to find the total probability of drawing two black shirts. Total Probability = (6/15) * (5/14) = (2/5) * (5/14) = 2/14 = 1/7.
7
When two unbiased coins are tossed, what is the probability of observing at most one head?
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Solution: Step 1: Determine the total number of possible outcomes when tossing two unbiased coins (Sample Space, S). S = {HH, HT, TH, TT} Total number of outcomes, n(S) = 4. Step 2: Identify the favorable outcomes where there is 'at most one head' (meaning zero or one head). E = {TT, HT, TH} Step 3: Count the number of favorable outcomes, n(E). n(E) = 3. Step 4: Calculate the probability of event E. P(E) = n(E) / n(S) = 3 / 4.
8
Under what condition is a binomial distribution considered to be negatively skewed?
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Solution: Step 1: Recall the relationship between the probability of success (p) and the skewness of a binomial distribution. Step 2: If p < 0.5 (i.e., p < 1/2), the distribution is positively (right) skewed. Step 3: If p = 0.5 (i.e., p = 1/2), the distribution is symmetrical. Step 4: If p > 0.5 (i.e., p > 1/2), the distribution is negatively (left) skewed. Step 5: Therefore, a binomial distribution is negatively skewed when 'p > 1/2'.
9
A basket contains 6 blue, 2 red, 4 green, and 3 yellow balls. If 5 balls are picked randomly, what is the probability that at least one is blue?
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Solution: Step 1: Calculate the total number of balls in the basket: 6 (blue) + 2 (red) + 4 (green) + 3 (yellow) = 15 balls. Step 2: Calculate the total number of ways to pick 5 balls from these 15 balls: n(S) = 15C5 = (15 × 14 × 13 × 12 × 11) / (5 × 4 × 3 × 2 × 1) = 3003. Step 3: Identify the complementary event E': None of the 5 balls picked are blue. Step 4: Count the number of non-blue balls: 2 (red) + 4 (green) + 3 (yellow) = 9 balls. Step 5: Calculate the number of ways to pick 5 balls from these 9 non-blue balls: n(E') = 9C5 = (9 × 8 × 7 × 6 × 5) / (5 × 4 × 3 × 2 × 1) = 126. Step 6: Calculate the probability of the complementary event P(E') = n(E') / n(S) = 126 / 3003 = 6 / 143. Step 7: The required probability P(E) = 1 - P(E') = 1 - (6/143) = (143 - 6) / 143 = 137 / 143.
10
An urn contains 6 red, 5 blue, and 2 green marbles. If two marbles are randomly selected, what is the probability that both selected marbles are red?
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Solution: Step 1: Calculate the total number of marbles in the urn: 6 + 5 + 2 = 13 marbles. Step 2: Calculate the total number of ways to pick 2 marbles from 13: n(S) = ¹³C₂ = (13 × 12) / (2 × 1) = 78. Step 3: Calculate the number of red marbles: 6. Step 4: Calculate the number of ways to pick 2 red marbles from 6 red marbles: n(E) = ⁶C₂ = (6 × 5) / (2 × 1) = 15. Step 5: Calculate the probability P(E) = n(E) / n(S) = 15 / 78. Step 6: Simplify the probability: 15 / 78 = 5 / 26.
11
What is the probability of drawing a face card from a standard deck of 52 playing cards?
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Solution: Step 1: Determine the total number of possible outcomes. A standard deck contains 52 cards. So, Total outcomes = 52. Step 2: Identify the number of favorable outcomes (face cards). In a standard deck, there are 4 suits (Hearts, Diamonds, Clubs, Spades). Each suit has 3 face cards: Jack, Queen, and King. Total number of face cards = 3 cards/suit \times 4 suits = 12 face cards. Step 3: Calculate the probability. Probability = (Favorable outcomes) / (Total outcomes) = 12 / 52. Step 4: Simplify the fraction. Divide both the numerator and denominator by their greatest common divisor, which is 4. 12 \div 4 = 3 52 \div 4 = 13 Simplified probability = 3 / 13.
12
When three fair coins are flipped, what is the probability of observing a maximum of two heads?
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Solution: Step 1: List all possible outcomes when three coins are tossed (sample space). S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} Total number of outcomes, n(S) = 2^3 = 8. Step 2: Define the event E as 'getting at most two heads'. This means 0 heads, 1 head, or 2 heads. E = {TTT, HTT, THT, TTH, HHT, HTH, THH} Number of favorable outcomes, n(E) = 7. Step 3: Calculate the probability. P(E) = n(E) / n(S) = 7 / 8.
13
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.
14
According to the special rule for adding probabilities, what kind of events are always involved?
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Solution: Step 1: Recall the special rule of addition for probability. Step 2: This rule states that the probability of A or B, P(A \cup B), is calculated as P(A) + P(B) if and only if events A and B cannot occur simultaneously. Step 3: Events that cannot occur simultaneously are defined as mutually exclusive events. Step 4: Therefore, in the special rule of addition of probability, the events are always mutually exclusive.
15
A school has 100 students. 60% are boys, and 40% of these boys play hockey. Girls do not play hockey, but 75% of girls play badminton. Only two games, hockey and badminton, are played in the school. How many students do not play any game?
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Solution: Step 1: Total students = 100. Step 2: Number of boys = 60% of 100 = 60. Step 3: Number of girls = 100 - 60 = 40. Step 4: Boys who play hockey = 40% of 60 = 24. Step 5: Girls who play badminton = 75% of 40 = 30. Step 6: Since girls do not play hockey, all 30 girls who play a game play badminton. The remaining girls (40 - 30 = 10) do not play badminton and do not play hockey, so 10 girls play no game. Step 7: For boys, we know 24 play hockey. The remaining boys are 60 - 24 = 36. Step 8: The problem states 'there are only two games to be played' (hockey and badminton). However, it does not specify how many of the remaining 36 boys play badminton, or if any of them play no game. Step 9: Without information on the remaining boys' participation in badminton (or lack thereof), it's impossible to determine the total number of students (boys + girls) who do not play any game. Step 10: Therefore, the number of students who don't play any game cannot be determined from the given information.
16
When two dice are tossed, what is the probability that their total score is a prime number?
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Solution: Step 1: Determine the total possible outcomes when two dice are thrown: n(S) = 6 × 6 = 36. Step 2: Identify the possible sums from two dice. The minimum sum is 1+1=2 and the maximum is 6+6=12. Step 3: List the prime numbers within this sum range: 2, 3, 5, 7, 11. Step 4: List the pairs of dice rolls that result in these prime sums: * Sum 2: (1, 1) - 1 way * Sum 3: (1, 2), (2, 1) - 2 ways * Sum 5: (1, 4), (2, 3), (3, 2), (4, 1) - 4 ways * Sum 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) - 6 ways * Sum 11: (5, 6), (6, 5) - 2 ways Step 5: Count the total number of favorable outcomes: n(E) = 1 + 2 + 4 + 6 + 2 = 15. Step 6: Calculate the probability P(E) = n(E) / n(S) = 15 / 36 = 5 / 12.
17
If you toss a coin and roll a die simultaneously, what is the probability of obtaining a tail on the coin and a 4 on the die?
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Solution: Step 1: Calculate the probability of getting a tail when tossing a coin: P(Tail) = 1/2. Step 2: Calculate the probability of getting a 4 when rolling a die: P(4) = 1/6. Step 3: Since the coin toss and die roll are independent events, multiply their individual probabilities to find the combined probability: P(Tail AND 4) = P(Tail) × P(4) = (1/2) × (1/6) = 1/12.
18
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.
19
Person A tells the truth in 75% of situations, and Person B tells the truth in 80% of situations. If they both narrate the same event, in what percentage of cases will their accounts contradict each other?
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Solution: Step 1: Define the probabilities of A and B telling the truth and lying. P(A_truth) = P(A) = 75% = 3/4 P(A_lie) = P(A') = 1 - 3/4 = 1/4 P(B_truth) = P(B) = 80% = 4/5 P(B_lie) = P(B') = 1 - 4/5 = 1/5 Step 2: Identify the scenarios where A and B contradict each other. Contradiction occurs if: a) A tells the truth AND B lies (P(A) × P(B')) b) A lies AND B tells the truth (P(A') × P(B)) Step 3: Calculate the probability of each contradiction scenario. P(A_truth and B_lie) = (3/4) × (1/5) = 3/20 P(A_lie and B_truth) = (1/4) × (4/5) = 4/20 = 1/5 Step 4: Since these two scenarios are mutually exclusive, add their probabilities to find the total probability of contradiction. P(Contradiction) = 3/20 + 4/20 = 7/20 Step 5: Convert the probability to a percentage. Percentage of contradiction = (7/20) × 100% = 35%.
20
What is the probability of obtaining a tail when a coin is tossed?
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Solution: Step 1: Identify all possible outcomes for a single coin toss (Sample Space, S). S = {Head (H), Tail (T)} Total number of outcomes, n(S) = 2 Step 2: Identify the favorable outcome. Favorable outcome = {Tail (T)} Number of favorable outcomes, n(E) = 1 Step 3: Calculate the probability. Probability (Tail) = n(E) / n(S) = 1/2
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