Question 1098384: A shipment of 22 fuses contains 3 defective fuses. A quality control specialist chooses a sample of 7 fuses from the shipment.
a. How many possible choices of 7 fuses can be made?
b. How many of these possible selections will not contain any defective fuses?
c. How many of the possible selections will contain at least one defective fuse?
d. How many of the possible selections will contain exactly one defective fuse?
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! A shipment of 22 fuses contains 3 defective fuses. A quality control specialist chooses a sample of 7 fuses from the shipment.
a. How many possible choices of 7 fuses can be made?
Combinations of 22, taking 7 at a time = 22x21x20x19x18x17x16 / (7x6x5x4x3x2) = 170544.
b. How many of these possible selections will not contain any defective fuses?
As there are 22 - 3 =18 fuses that are not defective, with those 18 you can make
18x17x16x15x14x13x12 / (7x6x5x4x3x2) = 50388
(combinations of 18, taking 7 at a time) sets of 7 fuses not containing any defective fuse.
c. How many of the possible selections will contain at least one defective fuse?
There are 170544 - 50388 = 120156
possible selections of 7 fuses (those not counted in part b above) that contain one or more defective fuses.
d. How many of the possible selections will contain exactly one defective fuse?
3 x 18x17x16x15x14x13x12 / (6x5x4x3x2) = 3 x 27132 = 81396 .
The number of possible sets of 6 not-defective fuses is
combinations of 18, taking 6 at a time = 18x17x16x15x14x13x12 / (6x5x4x3x2) = 27132.
To each of those you couls add one of the 3 defective fuses to make all possible selections containing exactly one defective fuse.
|
|
|
