Example Question - independent events
Here are examples of questions we've helped users solve.
Testing Independence of Events
Pour résoudre si les événements A et B sont indépendants, on utilise la définition de l'indépendance. Deux événements A et B sont indépendants si et seulement si P(A ∩ B) = P(A) * P(B). 1. Vérifions l'indépendance des événements A et B. On a : P(A) = 0.5, P(B) = 0.4, P(A ∩ B) = 0.2. Calculons P(A) * P(B) : P(A) * P(B) = 0.5 * 0.4 = 0.2. Puisque P(A ∩ B) = P(A) * P(B), les événements A et B sont indépendants dans ce cas. 2. Vérifions l'indépendance des événements A et B. On a: P(A) = 0.8, P(B) = 0.7, P(A ∩ B) = 0.65. Calculons P(A) * P(B) : P(A) * P(B) = 0.8 * 0.7 = 0.56. Puisque P(A ∩ B) ≠ P(A) * P(B), les événements A et B ne sont pas indépendants dans ce cas.
Calculating Probability of Independent Events
To solve this problem, we need to find the probability of two independent events: 1. Landing on a number less than 4. 2. Then landing on a prime number. Let's calculate them one by one: 1. Landing on a number less than 4: The spinner has numbers 1, 2, 3, 4, and 5. The numbers less than 4 are 1, 2, and 3. Hence, there are 3 favorable outcomes out of 5 possible outcomes. So, the probability of landing on a number less than 4 is 3/5. 2. Landing on a prime number: The prime numbers on the spinner are 2, 3, and 5. Therefore, there are 3 prime numbers out of 5 possible outcomes. So, the probability of landing on a prime number is also 3/5. Since both spins are independent events, the combined probability of both events occurring in succession is the product of their individual probabilities: Probability of both events = (Probability of first event) x (Probability of second event) = (3/5) x (3/5) = 9/25 To express this probability as a percentage, we convert the fraction to a decimal and then multiply by 100: 9/25 = 0.36 0.36 x 100 = 36% Therefore, the probability of landing on a number less than 4 and then landing on a prime number is 36%.
Calculating Probability of Sequential Events
To solve this problem, you need to calculate the probability of two independent events happening one after the other. The two events are: landing on an odd number first, and then landing on an even number. In the spinner shown in the image, there are 4 odd numbers (1, 3, 5, and 5) and 2 even numbers (2 and 4). The probability of landing on an odd number is the number of odd outcomes divided by the total number of outcomes. In this case: \( P(\text{odd number}) = \frac{\text{number of odd numbers}}{\text{total numbers on the spinner}} = \frac{4}{6} \) Since there are 2 even number outcomes, the probability of landing on an even number would be: \( P(\text{even number}) = \frac{\text{number of even numbers}}{\text{total numbers on the spinner}} = \frac{2}{6} \) To find the combined probability of landing on an odd number first and then on an even number, you multiply the probabilities of the two independent events: \( P(\text{odd then even}) = P(\text{odd number}) \times P(\text{even number}) \) Plugging in the values we have: \( P(\text{odd then even}) = \frac{4}{6} \times \frac{2}{6} = \frac{8}{36} \) You can simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 4: \( P(\text{odd then even}) = \frac{8 \div 4}{36 \div 4} = \frac{2}{9} \) Therefore, the simplified probability of spinning an odd number followed by an even number is \( \frac{2}{9} \).
Calculating Probability of Two Events Occurring in Succession
The question asks for the probability of two events happening in succession: rolling a prime number on a 6-sided die, and then rolling a number less than 4. The prime numbers on a 6-sided die are 2, 3, and 5. Therefore, there are 3 outcomes out of 6 that are prime numbers. The probability of rolling a prime number is therefore 3/6 or 1/2. The numbers less than 4 on a 6-sided die are 1, 2, and 3. So, there are 3 outcomes out of 6 that are less than 4. The probability of rolling a number less than 4 is also 3/6 or 1/2. Since these two events are independent (the outcome of the second roll does not depend on the outcome of the first roll), the total probability of both events occurring in succession is the product of their individual probabilities: (1/2) * (1/2) = 1/4 As a percentage, this is: 1/4 * 100 = 25% Therefore, the probability of rolling a prime number and then rolling a number less than 4 is 25%.
Calculating Probability of Independent Events
The problem is asking for the probability of two independent events: first picking a prime number, and then picking a number greater than 4. To solve this, we'll calculate the probability of each event and then multiply them together because the two events are independent. Prime numbers in the set provided are 2, 3, 5, and 7. There are a total of 8 distinct numbers. So, the probability of picking a prime number is: Number of prime numbers / Total numbers = 4/8 = 1/2 Numbers greater than 4 in the set are 5, 6, 7, and 8. Thus, the probability of then picking a number greater than 4 is again: Number of "greater than 4" numbers / Total numbers = 4/8 = 1/2 Since the events are independent, the overall probability is the product of the two individual probabilities: Probability of prime number AND number greater than 4 = (1/2) * (1/2) = 1/4 To express this as a percentage, we multiply by 100: 1/4 * 100 = 25% Rounded to the nearest tenth of a percent (although in this case there's no need to round since 25% is already at a tenth), the final answer is: 25.0%
Probability Calculation for Sequential Events
To solve this problem, we need to find the probability of two independent events occurring one after the other: picking a prime number and then picking a number greater than 4. There are 8 cards in total. Prime numbers in the set are 2, 3, 5, and 7. There are four prime numbers. Numbers greater than 4 in the set are 5, 6, 7, and 8. There are four numbers greater than 4. Since the events are independent (picking one card does not influence the next one, because the card is put back), we can multiply the probabilities of each event to get the total probability. The probability of picking a prime number is 4 out of 8 (since there are 4 prime numbers out of 8 total numbers). That gives us: P(prime) = 4/8 = 1/2 The probability of picking a number greater than 4 is also 4 out of 8 (since there are 4 numbers greater than 4 out of 8 total numbers). That gives us: P(greater than 4) = 4/8 = 1/2 Now, we multiply these probabilities together: P(prime and greater than 4) = P(prime) × P(greater than 4) P(prime and greater than 4) = 1/2 × 1/2 P(prime and greater than 4) = 1/4 As a percentage, we can calculate this by multiplying by 100: P(prime and greater than 4) = 1/4 × 100 = 25% So, the probability of picking a prime number and then picking a number greater than 4 is 25%. Since the question asks to round to the nearest tenth, our final answer remains 25.0%.
Calculating Probability of Independent Events
To solve this problem, we need to calculate the probability of two independent events occurring in sequence: 1. Picking a prime number card. 2. Picking a number greater than 4 card. The sample space consists of 8 cards, numbered from 1 to 8. Among these cards, the prime numbers are 2, 3, 5, and 7. Probability of picking a prime number card = number of prime cards / total number of cards P(prime) = 4/8 = 1/2 After picking a prime number, you replace the card and pick again. The number of cards greater than 4 are 5, 6, 7, and 8. Probability of picking a number greater than 4 = number of cards greater than 4 / total number of cards P(greater than 4) = 4/8 = 1/2 Since these are independent events, we can multiply the probabilities: P(prime and greater than 4) = P(prime) * P(greater than 4) P(prime and greater than 4) = (1/2) * (1/2) P(prime and greater than 4) = 1/4 Converting this into a percentage, we get: P(prime and greater than 4) = (1/4) * 100% = 25% Rounded to the nearest tenth of a percent, the answer remains 25.0%.
Calculating Probability of Events Happening Successively
To solve this probability question, we need to calculate the probability of two independent events happening one after the other: 1. Picking an even number. 2. Picking an 8. Since the card is put back after the first pick, the two events are independent, and we can calculate the probability of each event separately and then multiply them. First, let's find the probability of picking an even number from the cards shown. There are three cards, and only one of them is even (the 8). So the probability of picking an even number is: P(even) = Number of even cards / Total number of cards = 1/3 Next, let's find the probability of picking an 8. Since the card is put back after the first draw, the probability remains the same for picking an 8: P(8) = Number of cards with 8 / Total number of cards = 1/3 Now we multiply the probabilities of the two independent events: P(even, then 8) = P(even) * P(8) = (1/3) * (1/3) = 1/9 So, the probability of picking an even number and then picking an 8, with replacement, is 1/9.
Probability Calculation: Picking Even Number and 8
The problem in the image is asking us to calculate the probability of picking an even number and then picking an 8 when selecting from the cards labeled 7, 8, and 9. Firstly, let's determine the probability of picking an even number from the cards. Since there is only one even number (which is 8) out of three cards, the probability of picking an even number at the first try is 1/3. Since the card is put back after the first pick, the sample space does not change, and each pick is independent of the other. Next, we need to determine the probability of picking an 8. Well, because we replaced the card after the first pick, the probability of picking an 8 on the second pick is also 1/3. To find the overall probability of these two independent events happening one after the other (picking an even number and then picking an 8), we need to multiply the probabilities of each event occurring: Probability of picking an even number * Probability of picking an 8 = (1/3) * (1/3) = 1/9. So the probability of picking an even number and then picking an 8 in this scenario is 1/9.
Probability Question: Independent Events
To solve this probability question, you can use the fundamental counting principle. Since the events are independent (choosing a 6 and then a 5), you multiply the probabilities of each event happening separately. There are 8 cards, and there is one 6 in them. Once the 6 is picked, it is not replaced, so there are now 7 cards left. Among these remaining cards, there is one 5. The probability of picking a 6 on the first try is: P(6) = 1/8 Then, given that the 6 has been picked, there are 7 cards left, and the probability of picking a 5 is: P(5 after 6) = 1/7 Now multiply the two probabilities to find the overall probability of both events happening one after the other: P(6 then 5) = P(6) * P(5 after 6) = (1/8) * (1/7) = 1/56 To express this as a percentage, you divide 1 by 56 and then multiply by 100: Percentage = (1/56) * 100 ≈ 1.786% Rounded to the nearest tenth, it is 1.8%.
Probability Problem Involving Spinner
I can see that the image contains a question about a probability problem involving a spinner. The problem states that the spinner is divided into equally sized sections, 3 of which are gray and 5 of which are blue. The question asks for the probability that the first spin lands on gray and the second spin lands on blue. To solve this, we need to calculate the probability of both events happening one after the other. The chance that the first spin lands on gray is 3 out of 8, as there are 3 gray sections out of a total of 8. The probability of a specific event is calculated by the formula: \[ P(\text{Event}) = \frac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}} \] For the first spin: \[ P(\text{first spin on gray}) = \frac{3}{8} \] For the second spin, the probability that the spin lands on blue is 5 out of 8, as there are 5 blue sections. For the second spin: \[ P(\text{second spin on blue}) = \frac{5}{8} \] Since these are independent events, the overall probability of both occurring in sequence is the product of the two individual probabilities: \[ P(\text{gray then blue}) = P(\text{first spin on gray}) \times P(\text{second spin on blue}) \] \[ P(\text{gray then blue}) = \left( \frac{3}{8} \right) \times \left( \frac{5}{8} \right) \] \[ P(\text{gray then blue}) = \frac{3 \times 5}{8 \times 8} \] \[ P(\text{gray then blue}) = \frac{15}{64} \] So, the probability that the first spin lands on gray and the second spin lands on blue is \(\frac{15}{64}\).
Probability of Spinner Outcomes
The image shows three spinners, and the question asks what the probability is that the first spinner will land on A, the second spinner will land on an even number, and the third spinner will land on Blue. For Spinner 1, there appears to be two outcomes (A and B), so the probability that it will land on A is 1 out of 2. For Spinner 2, there are four outcomes (1, 2, 3, 4), and two of them are even (2 and 4), so the probability that it will land on an even number is 2 out of 4, which simplifies to 1 out of 2. For Spinner 3, there appears to be three outcomes (Blue, Red, and Green), so the probability that it will land on Blue is 1 out of 3. To find the overall probability of all three independent events happening, we multiply the probabilities together: (1/2) * (1/2) * (1/3) = 1/12. Therefore, the probability that the first spinner will land on A, the second spinner will land on an even number, and the third spinner will land on Blue is 1/12. The correct answer is option c.
Calculating Probability of Coin Flips
To solve this problem, we need to calculate the probability of flipping a coin twice and getting heads on the first flip and tails on the second flip. The probability of getting heads on a single flip is 1/2, and the probability of getting tails on a single flip is also 1/2. Since these two events are independent (the outcome of the first flip does not affect the outcome of the second flip), we can find the combined probability by multiplying the probabilities of the two individual events. So, the probability of getting heads first and then tails is: (1/2) * (1/2) = 1/4. Now to express this as a percentage, we convert the fraction to a percentage by multiplying by 100: 1/4 * 100% = 25%. Therefore, the probability of getting heads and then tails when flipping a coin twice is 25%.
Calculating Probability of Spinner Sequence
The image shows a question concerning probability from a spinner. The spinner has numbers 5, 6, and 7, and the task is to calculate the probability of first landing on a 6 and then on a 5 after spinning the spinner twice. To solve this, first calculate the individual probabilities of each separate event and then multiply them together since these are independent events. The probability of landing on a 6 on the first spin is: 1 out of 3, or 1/3, because there are three possible outcomes and only one of them is a 6. The probability of landing on a 5 on the second spin is: Also 1 out of 3, or 1/3 for the same reason. Now, to find the probability of both events occurring in sequence (landing on a 6 and then on a 5), multiply these two probabilities together: Probability of landing on a 6 and then a 5 = (1/3) * (1/3) = 1/9 Therefore, the probability of spinning a 6 first and then a 5 is 1/9.
Probability of Choosing Snacks in Melissa's Pantry
The image shows a table with the heading "The table shows the snacks in Melissa's pantry". There are three types of snacks listed in the table with their corresponding amounts: - Granola Bar: 4 - Chips: 3 - Fruit Snack: 5 The question asks: "What is the probability that she chooses a fruit snack, puts it back, and then chooses a granola bar?" To solve this, we calculate the probability for each event and then multiply them together because she is putting the snacks back after choosing, meaning the events are independent. Probability of choosing a fruit snack: The total number of snacks is 4 (Granola Bar) + 3 (Chips) + 5 (Fruit Snack) = 12 snacks. There are 5 fruit snacks, so the probability of choosing a fruit snack is 5/12. Since she puts the fruit snack back, the probabilities for the next choice remain the same. Probability of choosing a granola bar: The total number of snacks is still 12, and there are still 4 granola bars, so the probability of choosing a granola bar is 4/12, which simplifies to 1/3. Therefore, the probability of choosing a fruit snack and then a granola bar (independent events) is: (5/12) * (1/3) Now, let's calculate it: (5/12) * (1/3) = 5/36 So the probability that she chooses a fruit snack, puts it back, and then chooses a granola bar is 5/36.
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