AP EAMCET · Maths · Probability
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
- A \(\left(\frac{9}{10}\right)^3, \frac{1419}{1960}\)
- B \(\left(\frac{9}{10}\right)^2, \frac{1418}{1961}\)
- C \(\left(\frac{9}{10}\right)^2, \frac{1419}{1960}\)
- D \(\left(\frac{9}{10}\right)^3, \frac{1418}{1961}\)
Answer & Solution
Correct Answer
(A) \(\left(\frac{9}{10}\right)^3, \frac{1419}{1960}\)
Step-by-step Solution
Detailed explanation
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)…
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