A box contains 4 defective and 2 good parts.
Two parts are selected from the box (without replacement), and it
is recorded whether the
selected part is defective or good.
The probability that the first selected part is defective and the
second selected part is good is
__________
The probability that exactly one good part is selected is__________________
The probability that at least one defective part is selected is _____________________
Sol)
Given
A box contains 4 defective and 2 good parts
Total = 6 parts
Two are selected from the box
a)
Probability that first is defective and second is good
P ( first is defective ) = 4/6
As one is selected is defective in first order
P( second is good ) = 2/5
Total probability = 4C1/6C1 * 2C1/5C1
= 8/30
= 0.26666. = 0.27
b)
Probability that exactly on selected good part
P( one good is selected ) = 8/30 + 8/30
= 16/30
= 0.53333333 = 0.54
c)
Probability that atleast one defective part is selected
P( one is defective ) = (4C1 * 2C1 )/ 6C2
= 8 / 15
P( both are defective ) = 4C2 / 6C2
= 6/15
P(atleast one is defective ) = p( one ) + p( two )
= 8/15 + 6/15
= 14/15
= 0.9333333 = 0.94
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