Question

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. \( \frac{9}{10}^3 \times \frac{1419}{1960} \)
• B. \( \frac{9}{10}^2 \times \frac{1418}{1961} \)
• C. \( \frac{9}{10}^2 \times \frac{1419}{1960} \)
• D. \( \frac{9}{10}^3 \times \frac{1418}{1961} \)

Hint:

When calculating probabilities with and without replacement, ensure you adjust the denominator in each step for without replacement scenarios.

Answer 1

The Correct Option is A

We are given that: - Total number of screws: 50 - Defective screws: 5 - Non-defective screws: 50 - 5 = 45 % Option (a) With Replacement: When drawing with replacement, the probability that each screw drawn is non-defective is the same for each draw. The probability of drawing a non-defective screw in one trial is: \[ \frac{45}{50} = \frac{9}{10} \] Since there are 3 draws with replacement, the probability of drawing 3 non-defective screws is: \[ P(\text{all non-defective}) = \left( \frac{9}{10} \right)^3 \] Next, we need to calculate the probability of the event given that all the screws drawn are non-defective. This is: \[ P(\text{all non-defective}) = \left( \frac{9}{10} \right)^3 \times \frac{1419}{1960} \] % Option (b) Without Replacement: When drawing without replacement, the probability changes with each draw. We use the following formula: \[ P(\text{all non-defective}) = \frac{45}{50} \times \frac{44}{49} \times \frac{43}{48} \] This simplifies to: \[ P(\text{all non-defective}) = \frac{9}{10} \times \frac{1419}{1960} \] % Final Answer \[ \boxed{\frac{9}{10}^3 \times \frac{1419}{1960}} \]