Radio

A dept. store recieves shipment of 23 new radios. There are 5 defective in the shipment. If 4 radios are selected for display, what is the probability that 2 of them are defective?

The number of ways to select the two defective is:
5C2 = 5! / (2!3!)

The number of ways to select the two non-defective is:
18C2 = 18! / (2!16!).

The total number of ways to select 4 radios out of 23 is:
23C4 = 23! / (4!19!)

Multiply and divide to get the probability:
= (5C2)(18C2) / (23C4)
= (10)(153) / (8855)
= 306 / 1771 = .17278373800113

The answer is approximately .17278 or 17.3%

Good luck in your studies,
~ Mitch ~

One Response

1. Mitch Says:
   February 27th, 2010 at 9:17 pm

   The number of ways to select the two defective is:
   5C2 = 5! / (2!3!)

   The number of ways to select the two non-defective is:
   18C2 = 18! / (2!16!).

   The total number of ways to select 4 radios out of 23 is:
   23C4 = 23! / (4!19!)

   Multiply and divide to get the probability:
   = (5C2)(18C2) / (23C4)
   = (10)(153) / (8855)
   = 306 / 1771 = .17278373800113

   The answer is approximately .17278 or 17.3%

   Good luck in your studies,
   ~ Mitch ~
   References :

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