For this problem, assume the balls in the box are numbered 1 through 8, and that an experiment consists of randomly selecting 2 balls one after another without replacement.
What probability should be assigned to the event that at least one ball has
an odd number?
So this is the same as saying one odd ball and one even ball or two odd balls.
4/8 for first odd ball and 4/7 for even.
4/8 for first odd ball and 3/7 for second.
To find probability of both/all events occurring , multiply the probability of the individual events.
To find either-or probability, add the individual probabilities.
To find the probability of at least one ball having an odd number, we need to consider two scenarios:
1. Both balls have even numbers.
2. At least one ball has an odd number.
To calculate the probability of both balls having even numbers, we need to determine the probability of the first ball being even given that it is randomly selected without replacement. Since there are 4 out of 8 balls that are even, the probability of the first ball being even is 4/8.
After the first ball is selected, there are 7 balls left in the box, with 3 of them being even. Therefore, the probability of the second ball being even, given that the first ball was even, is 3/7.
To find the probability of both balls having even numbers, we multiply the probabilities of each ball being even:
(4/8) * (3/7) = 12/56 = 3/14.
Now, to find the probability of at least one ball having an odd number, we subtract the probability of both balls having even numbers from 1 (since the sum of the probabilities of all possible outcomes must be 1):
1 - (3/14) = 11/14.
Therefore, the probability assigned to the event that at least one ball has an odd number is 11/14.