In: Math
A box has 11 parts of which 4 are defective and 7 acceptable. 2 parts are chosen at random without replacement. Find the probability that:
a) both parts are defective.
b) both parts are acceptable.
c) only one part is defective.
Box has 4 defective parts and 7 acceptable parts
a) Probability of picking a defective part in first pick = 4/11
Probability of picking a defective part in second pick = 3/10 since 1 defective part was already picked
Therefore probability of picking both defective parts = (4/11)*(3/10) = 6/55 = 0.1091
b) Probability of picking an acceptable part in first pick = 7/11
Probability of picking an acceptable part in second pick = 6/10 since 1 acceptable part was already picked
Therefore probability of picking both acceptable parts = (7/11)*(6/10) = 21/55 = 0.3818
c) Case 1:
Probability of picking an acceptable part in first pick = 7/11
Probability of picking a defective part in second pick = 4/10
Case 2:
Probability of picking a defective part in first pick = 4/11
Probability of picking an acceptable part in second pick = 7/10
Therefore probability of picking only one defective part = (7/11)*(4/10) +(4/11)*(7/10) = 28/55 = 0.5091