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Three screws are drawn at random from a lot of 50 screws containing 5 defective ones. Then the probability of the event that all 3 screws drawn are non-defective, assuming that the drawing is (a) with replacement (b) without replacement respectively is
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Verified Answer
The correct answer is:
$\left(\frac{9}{10}\right)^3, \frac{1419}{1960}$
Given total screws $=50$
defective screws $=5$
and non-defective screws $=45$
Let $\mathrm{A}$ be the event of getting drawing of 3 screws are not defective.
(a) with replacement,
$P(A)=\left(\frac{{ }^{45} C_1}{{ }^{50} C_1}\right)^3=\left(\frac{9}{10}\right)^3$
(b) without replacement
$\begin{aligned}
& P(A)=\frac{{ }^{45} C_1}{{ }^{50} C_1} \times \frac{{ }^{44} C_1}{{ }^{49} C_1} \times \frac{{ }^{43} C_1}{{ }^{48} C_1} \\
& =\frac{45}{50} \times \frac{44}{49} \times \frac{43}{48}=\frac{1419}{1960}
\end{aligned}$
defective screws $=5$
and non-defective screws $=45$
Let $\mathrm{A}$ be the event of getting drawing of 3 screws are not defective.
(a) with replacement,
$P(A)=\left(\frac{{ }^{45} C_1}{{ }^{50} C_1}\right)^3=\left(\frac{9}{10}\right)^3$
(b) without replacement
$\begin{aligned}
& P(A)=\frac{{ }^{45} C_1}{{ }^{50} C_1} \times \frac{{ }^{44} C_1}{{ }^{49} C_1} \times \frac{{ }^{43} C_1}{{ }^{48} C_1} \\
& =\frac{45}{50} \times \frac{44}{49} \times \frac{43}{48}=\frac{1419}{1960}
\end{aligned}$
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