the numbers 1 through 9 are written on nine slips of paper and placed into a hat. If slips are selected from the hat one at a time at random, what is the minimum number that must be chosen to ensure that the sum in these slips exceeds 10
To find the minimum number of slips that must be chosen to ensure that the sum exceeds 10, we need to consider the worst-case scenario.
First, let's determine the sum of all the numbers from 1 to 9. We can do this by using the formula for the sum of an arithmetic series: S = (n/2)(first term + last term), where S is the sum, n is the number of terms, and the first and last terms are 1 and 9, respectively.
S = (9/2)(1 + 9) = (9/2)(10) = 45
Since the sum of all the numbers is 45, if we choose the first 9 slips without getting a sum higher than 10, there is a possibility that the sum will be exactly 10. Therefore, we need to choose at least one more slip to ensure that the sum exceeds 10.
To calculate the sum of the first 9 slips, we can again use the formula for the sum of an arithmetic series:
S = (n/2)(first term + last term) = (9/2)(1 + 9) = 45/2 = 22.5
Since we cannot choose a fractional slip, we need to round up to the nearest whole number. Therefore, we need to choose at least 23 slips to ensure that the sum exceeds 10.
Thus, the minimum number of slips that must be chosen to ensure that the sum exceeds 10 is 23.
How bad could it get?
1+2+3+4 = 10
so at least five slips to be sure to get a sum greater than ten.