Question

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 key boards 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.
Round your answers to four decimal places.
a. What is the probability that this shipment will be accepted?
P(shipment accepted)=
b. What is the probability that this shipment will not be accepted?
P(shipment not accepted)=

Answer 1

Answer:

a

  $P(shipment\ accepted) =0.6337$\

b

   $P(shipment\ not\ accepted)= 0.3663$  

Step-by-step explanation:

from the question we are told that

   The number of boxes N  =  100

     The  number of keyboard in each box is  k =  20  

     The sample size is  n =  5

Generally the number of ways of selecting 1 keyboard from the 5 selected keyboard is mathematically represented as

     $G = \ ^{5}C_1$

Here  C stands for  combination (Hence in the question we will be making use of the combination functionality in our calculator )

Generally the number of ways of selecting 0 keyboard from the 5 selected keyboard is mathematically represented as

       $H = \ ^{5}C_0$

Generally the number of ways of selecting 5 keyboard from the 20 keyboards is mathematically represented as

          $F = \ ^{20}C_5$

Generally the number of ways of selecting 5 keyboard from the  15 keyboards is mathematically represented as

        $W = \ ^{15}C_5$

Generally the number of ways of selecting 5 keyboard from the  15 keyboards is mathematically represented as

        $V = \ ^{15}C_4$

Generally the probability that a shipment will be accepted is mathematically represented as

 Probability of  0 defect out of  5 + Probability of  1 defect out of 5

Now  

  Probability of  0 defect out of  5 is mathematically represented  as

        $P(A_1) = \frac{H*W }{ F}$

=>      $P(A_1) = \frac{1 *3003 }{ 15504}$

=>      $P(A_1) = 0.1937$

And  Probability of  1 defect out of 5 is mathematically represented as

     $P(A_2) = \frac{ G * V}{ F}$

 => $P(A_2) = \frac{ 5 * 1365}{15504}$  

 =>   $P(A_2) = 0.440$  

Generally the probability that a shipment will be accepted is mathematically represented as

     $P(shipment\ accepted) =0.1937 + 0.440$

=>    $P(shipment\ accepted) =0.6337$

Generally the probability that this shipment will not be accepted is mathematically represented as

    $P(shipment\ not\ accepted)= 1 -P(shipment \ accepted)$

     $P(shipment\ not\ accepted)= 1 - 0.6337$  

     $P(shipment\ not\ accepted)= 0.3663$