Mansoor Tuesday, December 16, 2014 at 8:44pm
Bender Electronics buys keyboards for its computers from another company. The keyboards are received
in shipments of 100 boxes, each box containing 20 keyboards. The quality control department at
Bender Electronics first randomly selects one box from each shipment and then randomly selects 5 keyboards
from that box. The shipment is accepted if not more than 1 of the 5 keyboards is defective. The
quality control inspector at Bender Electronics selected a box from a recently received shipment of keyboards.
Unknown to the inspector, this box contains 6 defective keyboards.
a. What is the probability that this shipment will be accepted?
b. What is the probability that this shipment will not be accepted?
for part a is 0.516 and for part b is 0.484
To find the probability that the shipment will be accepted, we need to calculate the probability of selecting 5 non-defective keyboards from the box.
a. What is the probability that this shipment will be accepted?
To calculate this probability, we need to use the concept of combinations. The total number of combinations of selecting 5 keyboards from a box of 20 is given by:
C(20, 5) = 20! / (5!(20-5)!)
= 20! / (5!15!)
= (20 * 19 * 18 * 17 * 16) / (5 * 4 * 3 * 2 * 1)
= 15,504
Now, let's calculate the probability of selecting 5 non-defective keyboards. Since there are 6 defective keyboards in the box, the number of non-defective keyboards is 20 - 6 = 14.
The total number of combinations of selecting 5 non-defective keyboards from a box of 14 is given by:
C(14, 5) = 14! / (5!(14-5)!)
= 14! / (5!9!)
= (14 * 13 * 12 * 11 * 10) / (5 * 4 * 3 * 2 * 1)
= 2,772
Therefore, the probability of selecting 5 non-defective keyboards is 2,772 / 15,504 = 0.1786, or approximately 17.86%.
Since the shipment is accepted if not more than 1 of the 5 keyboards is defective, we also need to consider the probability of selecting exactly 1 defective keyboard.
The number of combinations of selecting 1 defective keyboard from the 6 in the box and 4 non-defective keyboards from the 14 remaining non-defective keyboards is given by:
C(6, 1) * C(14, 4) = (6! / (1!(6-1)!)) * (14! / (4!(14-4)!))
= (6 * 14 * 13 * 12 * 11) / (1 * 4 * 3 * 2 * 1)
= 8,355
Therefore, the probability of selecting exactly 1 defective keyboard and 4 non-defective keyboards is 8,355 / 15,504 = 0.5384, or approximately 53.84%.
The probability of the shipment being accepted is the sum of the probabilities of selecting 5 non-defective keyboards and exactly 1 defective keyboard:
0.1786 + 0.5384 = 0.717, or approximately 71.7%.
Therefore, the probability that this shipment will be accepted is 0.717, or approximately 71.7%.
b. What is the probability that this shipment will not be accepted?
To find the probability that the shipment will not be accepted, we simply subtract the probability of accepting the shipment from 1:
1 - 0.717 = 0.283, or approximately 28.3%.
Therefore, the probability that this shipment will not be accepted is 0.283, or approximately 28.3%.