A box contains 7 b
alls marked with numbers 1, 2, 3, 4, 5, 6, and 7. We randomly select 2 balls from the box without replacement. What is the probability that at least one of the balls has an odd number?
There are a total of 7 balls in the box, and we need to select 2 of them. The total number of ways we can do this is given by the combination formula:
C(7,2) = 7! / (2! (7-2)!) = 21
Now we need to find the number of ways we can select 2 balls such that at least one of them has an odd number. There are two cases to consider:
Case 1: One ball is odd, and the other is even.
There are 4 odd balls and 3 even balls in the box. We can choose one odd ball in C(4,1) ways and one even ball in C(3,1) ways. The total number of ways to select one odd and one even ball is:
C(4,1) * C(3,1) = 12
Case 2: Both balls are odd.
There are 4 odd balls in the box, so we can choose two odd balls in C(4,2) ways:
C(4,2) = 6
The total number of ways to select 2 balls with at least one odd number is the sum of the number of ways in Case 1 and Case 2:
12 + 6 = 18
Therefore, the probability of selecting 2 balls with at least one odd number is:
18 / 21 = 6 / 7
Answer: The probability that at least one of the balls has an odd number is 6/7.