Question

In a factory, it is known that the probability of fixing a particular production error is 1/3. For production of 10 units with this error;
a-)Write the probability function.
b-)What is the probability that all three can be corrected?
c-)What is the probability that one of the most errors can be fixed?
d-)What is the probability that two or more errors can be corrected?
e-)What is the probability that more than three, 5 or fewer errors can be corrected?

Answer 1

Let x be the number of units fixed or corrected.

x follows binomial distribution with n = 10, p = 1/3 and q = 1-(1/3) = 2/3

a) probability function :

P( X = x ) = ; x = 0,1,2,…n

b)What is the probability that all three can be corrected?

P( x = 3 )

=

= 120*0.0370*0.0585

= 0.2601

c) What is the probability that one of the most errors can be fixed?

P( x = 1 )

=

= 10*0.3333*0.0260

= 0.0867

d)What is the probability that two or more errors can be corrected?

P( x ≥ 2 ) = 1 - P ( x ≤ 1 ) = 1 - P( x = 0 ) - P( x = 1 )

P( x = 0 ) =

= 1* 1 * 0.0173

= 0.0173

P ( x = 1) = 0.0867

P( x ≥ 2 ) = 1 - 0.0173 - 0.0867

P( x ≥ 2 ) = 0.8960

e)What is the probability that more than three, 5 or fewer errors can be corrected?

P( 3 < x ≤ 5 ) = P( 4 ≤ x < 5 )

P( x = 4 ) =

= 210*0.0123*0.0878

= 0.2276

P( x = 5 ) =

= 252*0.0041*0.1317

= 0.1366

P( 3 < x ≤ 5 ) = 0.2276 +0.1366

P( 3 < x ≤ 5 ) = 0.3642