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(A) Total screws $= 50$.Defective screws $= 5$.Non-defective screws $= 45$.Let $A$ be the event that all $3$ screws drawn are non-defective.$(a)$ With

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$\left(\frac{9}{10}\right)^3, \frac{1419}{1960}$

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(A) Total screws $= 50$.Defective screws $= 5$.Non-defective screws $= 45$.Let $A$ be the event that all $3$ screws drawn are non-defective.$(a)$ With replacement:The probability of drawing a non-defective screw in one draw is $P = \frac{45}{50} = \frac{9}{10}$.Since the drawing is with replacement,the events are independent.$P(A) = \left(\frac{9}{10}ight) 	imes \left(\frac{9}{10}ight) 	imes \left(\frac{9}{10}ight) = \left(\frac{9}{10}ight)^3$.$(b)$ Without replacement:The probability is calculated as:$P(A) = \frac{45}{50} 	imes \frac{44}{49} 	imes \frac{43}{48}$.$P(A) = \frac{9}{10} 	imes \frac{44}{49} 	imes \frac{43}{48} = \frac{3}{10} 	imes \frac{11}{49} 	imes \frac{43}{4} = \frac{1419}{1960}$.

Three screws are drawn at random from a lot of $50$ screws containing $5$ defective ones. The probability of the event that all $3$ screws drawn are non-defective,assuming that the drawing is $(a)$ with replacement and $(b)$ without replacement,is respectively:

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