Originally Posted by

**s_ingram** 3 people independently think of an integer in the set {1,2,3,4,5,6,7}. Find in fractional form the probability that:

a) all 3 of the integers selected are greater than 4

b) the least integer selected is 5

c) the 3 integers selected are different, given that the least integer selected is 5

d) the sum of the 3 integers selected is more than 15

so this is a problem with replacement.

a) The probability of selecting a number greater than 4 is 3/7 and there are three numbers to select so the probability is $(3/7)^3 = \frac{27}{343}$

b) if the least number selected is 5, there are 3 numbers 5, 6 and 7 that can be arranged in 27 ways. We require only those selections that contain a 5.

an x in column 4 identifies all the entries containing a 5. There are 19 'x's, so the Probability is $\frac{19}{343}$

c) now we have to fish out the number of cases where all 3 numbers are different. Again the table is handy and I wouldn't know how to proceed without it.

These are marked with a y in Column 5 and there are 6 values.

d) to get the sum of the integers more than 15, it is just a case of writing out all the possible numbers and counting the cases. I won't write it all out here.

As I said, my suspicion is that there its a better way!