Example Question - sequential events
Here are examples of questions we've helped users solve.
Calculating Probability of Sequential Events
To solve this probability question, we need to calculate the probability of two events happening in sequence: picking a card with the number 1 and then picking an even-numbered card after that without replacing the first card. The first event is picking a 1. There are three cards, and only one of them is a 1, so the probability of picking a 1 is: P(1) = 1/3 Now, if the 1 card has been picked, there are only two cards remaining. To satisfy the second event, we need to pick an even number from the two remaining cards. There is only one even card left (either 2 or 3 was removed, depending on which was picked), so the probability of picking an even number after picking 1 is: P(even | 1) = 1/2 To find the overall probability of both events occurring, we multiply the probabilities of the two events: P(1 and then even) = P(1) * P(even | 1) = (1/3) * (1/2) = 1/6 The probability of picking a 1 and then picking an even number in sequence without replacement is 1/6.
Calculating Dependent Probability of Sequential Events
To solve this problem, we need to calculate the probability of two events happening one after the other without replacement, which is a dependent probability scenario. The two events are: 1. Picking a card with the number 1 on it. 2. Picking a card with an even number on it (without replacing the first card). First, let's calculate the probability of the first event: There are 3 cards, and only one card has the number 1 on it. So, the probability of picking the number 1 is: P(1) = 1/3 Now, assuming you picked the card with the number 1 on it, there are now 2 cards left. The second event is picking an even number, and among the remaining 2 cards (since the first card is not replaced), there is only one even number (which is number 2). So, the probability of picking an even number after picking number 1 is: P(even | 1) = 1/2 To find the overall probability of both events happening in succession, we multiply the probabilities of each individual event: P(1 and then an even) = P(1) * P(even | 1) = (1/3) * (1/2) = 1/6 Therefore, the probability of picking a 1 and then picking an even number is 1/6.
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%.
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