Option A
is one way to determine the factors of
by grouping
<em><u>Solution:</u></em>
Factoring by grouping means that you will group terms with common factors before factoring
<em><u>Given expression is:</u></em>
Group the first two terms together and then the last two terms together.
We can see that is common in first two terms
And 3 is common in last two terms
Factor them out
Thus option A is correct
Answer:
a) P (0 defective component) = 0.5
P( 1 defective component ) = 0.35
P( 2 defective component ) = 0.15
b) P( 0 ) = 0
p ( 1 ) = 1
p ( 2 ) = 0.33
Step-by-step explanation:
p( no defective ) = 0.5
p( 1 is defective ) = 0.35
p( 2 is defective ) = 0.15
Given that 2 components are selected at random
<u>a) Given that neither component is defective </u>
Probability of 0 defective component = 0.5
P( 1 defective component ) = 0.35
P( 2 defective component ) = 0.15
<u>b) Given that one of the two tested component is defective </u>
P( 0 defective ) = 0
P( 1 defective ) = P ( ) = p( x = 1 ) / 1 - P ( x = 0 )
= ( 0.5 )^1 ( 0.5 )^0 / 1 - ( 0.5)^0 (0.5)^1
= 0.5 * 1 / 1 - 0.5 = 0.5 / 0.5 = 1
p ( 2 defective ) = p( x = 3 ) / 1 - P ( x = 0 )
= ( 0.5 )^2 ( 0.5 )^0 / 1 - ( 0.5)^0 (0.5)^2
= 0.25 / 0.75 = 0.33
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