Answer:
a) 0.00019923%
b) 47.28%
Step-by-step explanation:
a) To find the probability of all sockets in the sample being defective, we can do the following:
The first socket will be in a group where 5 of the 38 sockets are defective, so the probability is 5/38
The second socket will be in a group where 4 of the 37 sockets are defective, as the first one picked is already defective, so the probability is 4/37
Expanding this, we have that the probability of having all 5 sockets defective is: (5/38)*(4/37)*(3/36)*(2/35)*(1/34) = 0.0000019923 = 0.00019923%
b) Following the same logic of (a), the first socket have a chance of 33/38 of not being defective, as we will pick it from a group where 33 of the 38 sockets are not defective. The second socket will have a chance of 32/37, and so on.
The probability will be (33/38)*(32/37)*(31/36)*(30/35)*(29/34) = 0.4728 = 47.28%
