The spinner is divided into 6 equal parts. If it is spun twice, find the probability that
a. Both spins stop on even numbers
b. Both spins stop on odd numbers.
c. One is odd and the other is even
d. The sum on the first spin and second spin is less than or equal to 8.
e. The sum of the first spin and second spin is more than 8
Please help me to answer this
Hhj
ST.LOUIS COLLEGE
To find the probability for each scenario, we need to understand the total possible outcomes and the favorable outcomes for each case.
a. Both spins stop on even numbers:
There are three even numbers (2, 4, 6) on the spinner. Since each spin is independent, the probability of the first spin stopping on an even number is 3/6, and the probability of the second spin also stopping on an even number is also 3/6. To find the probability of both events occurring, we multiply the probabilities together: (3/6) * (3/6) = 9/36 = 1/4.
b. Both spins stop on odd numbers:
Similar to the previous case, there are three odd numbers (1, 3, 5) on the spinner. The probability of the first spin stopping on an odd number is 3/6, and the probability of the second spin also stopping on an odd number is also 3/6. Therefore, the probability of both spins stopping on odd numbers is (3/6) * (3/6) = 9/36 = 1/4.
c. One is odd and the other is even:
In this case, we need to consider the scenarios where one spin stops on an odd number and the other stops on an even number. There are three odd numbers and three even numbers on the spinner, so the probability of the first spin stopping on an odd number is 3/6, and the probability of the second spin stopping on an even number is also 3/6. However, we need to consider the two possible orders: odd-even and even-odd. Therefore, we need to multiply the calculated probability by 2. Thus, the probability of one spin stopping on an odd number and the other spin stopping on an even number is 2 * (3/6) * (3/6) = 2 * 9/36 = 18/36 = 1/2.
d. The sum on the first spin and second spin is less than or equal to 8:
To find the favorable outcomes for this case, we need to consider the possible combinations of numbers that sum up to less than or equal to 8. The favorable combinations are:
(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)
There are a total of 36 possible outcomes (6 numbers on the spinner for each spin). Therefore, the probability is 10/36 = 5/18.
e. The sum of the first spin and second spin is more than 8:
To find the favorable outcomes for this case, we need to consider the possible combinations of numbers that sum up to more than 8. The favorable combinations are:
(3, 6), (4, 5), (5, 4), (6, 3), (6, 5), (5, 6)
Therefore, there are 6 favorable outcomes out of a total of 36 possible outcomes, resulting in a probability of 6/36 = 1/6.