Question 1081308
a. That would be 4/11 with a uniform distribution with each weight between 95 and 106 being equally likely.
b. Probability of 1 of them is 5*(4/11)(7/11)^4=0.2982
Probability of 0 of them is (7/11)^5=0.1044.  That adds to 0.4026, and at least 2 would be the complement or 0.5974.
c. It is between 102 and 110 is really between 102 and 106 or 4/11.
d. 101 gm is 1/11.