Question

Q‒2. [4×5 marks] A box contains 24 transistors, 4 of which are defective. If
4 are sold at random, find the following probabilities.
a) Exactly 2 are defectives.
b) None is defective.
c) All are defective.
d) At least 1 is defective.

Find the transitive closure of if is
a) .
b) .

Answer 1

This is an example of Hypergeometric distribution with following parameters :
N = 24, M = 4, N-M = 20, n = 4
where N represents the total number of transistors in the population , M represents the total number of defective transistors in the population and n represents the randomly selected sample of transistors from N.

In general,
P ( X = x ) = [ MCx * N-MCn-x ] / NCn

a) P( Exactly 2 are defectives ) = P( X = 2 )
= [ 4C2 * 20C4-2 ] / 24C4
= 0.107284

b) P( None is defective.) = P( X = 0 )
= [ 4C0 * 20C4-0 ] / 24C4
   = 0.4559571

c) P( All are defective.) = P( X = 4 )
= [ 4C4 * 20C4-4 ] / 24C4
   = 0.00009412

d) P ( Atleast 1 is defective ) = P ( X >= 1 )
= 1 - P ( X = 0 )
= 1 - 0.4559571
= 0.5440429

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