Answer:
a) 0.00019923%
b) 47.28%
Step-by-step explanation:
a) To determine the likelihood that all sockets in the sample are defective, we can use the following approach:
The first socket is among a group that has 5 defective out of 38, leading to a probability of 5/38.
The second socket is then taken from a group of 4 defective out of 37, following the selection of the first defective socket, resulting in a probability of 4/37.
Extending this logic, the chance of having all 5 defective sockets is computed as: (5/38)*(4/37)*(3/36)*(2/35)*(1/34) = 0.0000019923 = 0.00019923%.
b) Using similar reasoning as in part a, the first socket has a probability of 33/38 of not being defective as it's chosen from a set where 33 sockets are functionally sound. The next socket has a proportion of 32/37, and this continues onward.
The overall probability calculates to (33/38)*(32/37)*(31/36)*(30/35)*(29/34) = 0.4728 = 47.28%.
