Example Question - probability of sequential events
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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 \( \frac{2}{9} \).
Probability of Sequential Events
To solve this problem, we need to use the concept of probability and calculate the chance of two independent events happening in succession. The spinner has 5 sections marked with numbers 1 through 5. First, we determine the probability of landing on an odd number. The odd numbers on the spinner are 1, 3, and 5. There are three odd sections out of a total of 5 sections. Probability of landing on an odd number: \[ P(\text{odd}) = \frac{\text{Number of odd sections}}{\text{Total number of sections}} = \frac{3}{5} \] Next, we determine the probability of landing on an even number. The even numbers on the spinner are 2 and 4. There are two even sections out of a total of 5 sections. Probability of landing on an even number: \[ P(\text{even}) = \frac{\text{Number of even sections}}{\text{Total number of sections}} = \frac{2}{5} \] Since we're dealing with two independent events (the result of the first spin doesn't affect the second spin), we multiply the two probabilities to find the overall probability of both events occurring in sequence. \[ P(\text{odd then even}) = P(\text{odd}) \times P(\text{even}) = \frac{3}{5} \times \frac{2}{5} = \frac{3 \times 2}{5 \times 5} = \frac{6}{25} \] Thus, the probability of landing on an odd number and then on an even number is \(\frac{6}{25}\).
Probability of Sequential Events
To solve this probability question, we must consider that the events are sequential and assume that, unless stated otherwise, a chosen number is not replaced. The first event is picking a 6 and the second event is picking a 5. The probability of picking a 6 first: There are 8 numbers in total. Only one of these numbers is a 6. So, the probability of picking a 6 is 1 out of 8, or: \[ P(6) = \frac{1}{8} \] Assuming the 6 is not replaced, we now have 7 numbers left. The probability of picking a 5 next: Since one number (the 6) has already been picked and is not being replaced, there are now 7 numbers remaining. Again, only one of these numbers is a 5. So, the probability of picking a 5 next is 1 out of 7, or: \[ P(5) = \frac{1}{7} \] To find the overall probability of both events happening in sequence (picking a 6 and then a 5), you multiply the probabilities of each individual event: \[ P(6 \text{ and then } 5) = P(6) \times P(5) = \frac{1}{8} \times \frac{1}{7} \] \[ P(6 \text{ and then } 5) = \frac{1}{56} \] Therefore, the probability of picking a 6 and then picking a 5 is \( \frac{1}{56} \).
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