SOLUTION: A shipment of assembly parts from a vendor offering inexpensive parts is used in a manufacturing plant. The box of 12 parts contains 5 that are defective and will not fit during as

Algebra ->  Probability-and-statistics -> SOLUTION: A shipment of assembly parts from a vendor offering inexpensive parts is used in a manufacturing plant. The box of 12 parts contains 5 that are defective and will not fit during as      Log On


   

Question 1181523: A shipment of assembly parts from a vendor offering inexpensive parts is used in a manufacturing plant. The box of 12 parts contains 5 that are defective and will not fit during assembly. A worker picks parts one at a time and attempts to install them. Find the probability of each outcome.
(a) The first two chosen are both good.
(b) At least one of the first three is good.
(c) The first four picked are all good.
(d) The worker has to pick five parts to find one that is good.
Answer by ikleyn(52786) About Me  (Show Source):
You can put this solution on YOUR website!
.
A shipment of assembly parts from a vendor offering inexpensive parts is used in a manufacturing plant.
The box of 12 parts contains 5 that are defective and will not fit during assembly.
A worker picks parts one at a time and attempts to install them. Find the probability of each outcome.
(a) The first two chosen are both good.
(b) At least one of the first three is good.
(c) The first four picked are all good.
(d) The worker has to pick five parts to find one that is good.
~~~~~~~~~~~~~~


At the beginning, there are 7 good and 5 defective parts in the set of 12 parts. 


(a)  P = %287%2F12%29%2A%286%2F11%29 = reduce fractions = %287%2F2%29%2A%281%2F11%29 = 7%2F22.    ANSWER


         Quite simple:  7%2F12  is the probability to get first good part from the set of 12 parts;

                        6%2F11  is the probability to get the second good part from remaining 6 good 
                              parts among remaining 11 parts.
     


(b)  P =  = %287%2A5+%2B+28%2A5+%2B+35%2A1%29%2F220 = 210%2F220 =  is the total number of different selection 3 parts of 12 parts.

         The numerator is the number of favorable sets of 3 parts, containing at least 1 good part of the 3.



(c)  P = %287%2F12%29%2A%286%2F11%29%2A%285%2F10%29%2A%284%2F9%29 = reduce fractions = %287%2F2%29%2A%281%2F11%29%2A%281%2F2%29%2A%284%2F9%29 = 7%2A%281%2F11%29%2A%281%2F9%29 = 7%2F99.    ANSWER


         The same logic works as in part (a).

Let me stop at this point: I do not like to answer tooooooooo many questions at a a time.
It is not a good style of teaching.