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The probability of a manufacturing company producing a defective item is 0.1. If 5 items are drawn at random from a set of 100 distinct items, what is the probability that at least 2 items would be defective?
  • a)
    1-0.94 x 1.4
  • b)
    1-0.94
  • c)
    1-0.95
  • d)
    1-0.94 x 0.5
  • e)
    1-0.94 x 0.1
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
The probability of a manufacturing company producing a defective item ...
Given
  • P(defective item) = 0.1
    • P(non-defective item) = 0.9
  • Total number of items = 100
  • Number of items drawn = 5
To Find: Probability that at-least 2 items will be defective
Approach
  1. Event= getting (2, 3, 4, or 5) items defective
  2. Non-Event = getting (0 or 1) item defective
    1. So, we will opt for the non-event method
  3. P(Event) = 1 – P(non-event)
  4. Non- Event: Getting 0 or 1 item defective
  1. Case-I: Getting 0 defective items
  2. Total number of ways in which 0 selected items can be defective = Item-1 is non-defective AND item-2 is non-defective…………AND item-5 is non defective
  3. = P(item-1 being non-defective) * P(item-2 being non-defective) *P(item-3 being non-defective)……….* P(item-5 being non-defective)
  4. Case-II: Getting 1 defective item
  5. P(getting 1 item defective AND 4 items non-defective) = Number of ways of selecting 1 item from 5 items * P(an item being defective) * P(an item being non-defective) *…..P(item being non-defective)
  6. P(Non-Event) = P(case-I) + P(case-II)
 
Working Out
  1. Non-Event: getting 0 or 1 defective items
  1. Case-I: Getting 0 defective items
  2. P(getting 0 defective items) = P(item-1 being non-defective) *P(item-2 being non-defective) …….P(item-5 being non-defective) =0.95…….(1)
  3. Case-II: getting 1 defective item
  4. P(getting 1 item defective AND 4 items non-defective) = Number of ways of selecting 1 item from 5 items * P(an item being defective) * P(an item being non-defective) *…..P(item being non-defective)
  5. P(Non-event-) = 
  6. 2.  P(Event) = 1- P(non-Event) =
 
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Most Upvoted Answer
The probability of a manufacturing company producing a defective item ...
To solve this problem, we can use the concept of binomial distribution. Let's break down the problem into smaller steps to find the probability of at least 2 defective items.

Step 1: Finding the probability of exactly 2 defective items
The probability of producing a defective item is given as 0.1. Therefore, the probability of producing a non-defective item is 1 - 0.1 = 0.9.
We can use the binomial probability formula to find the probability of exactly 2 defective items out of 5 drawn items:
P(X = 2) = (5C2) * (0.1^2) * (0.9^3)
= (5! / (2! * 3!)) * (0.1^2) * (0.9^3)
= (10) * (0.01) * (0.729)
= 0.0729

Step 2: Finding the probability of exactly 3 defective items
Using the same formula, we can find the probability of exactly 3 defective items out of 5 drawn items:
P(X = 3) = (5C3) * (0.1^3) * (0.9^2)
= (5! / (3! * 2!)) * (0.001) * (0.81)
= 0.0729

Step 3: Finding the probability of at least 2 defective items
To find the probability of at least 2 defective items, we need to find the sum of the probabilities of exactly 2, 3, 4, and 5 defective items.
P(at least 2) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

Since the number of defective items can only range from 0 to 5, we know that P(X = 4) and P(X = 5) will be very small probabilities, approaching zero. Therefore, we can ignore them in our calculation.
P(at least 2) ≈ P(X = 2) + P(X = 3)

P(at least 2) ≈ 0.0729 + 0.0729
P(at least 2) ≈ 0.1458

Step 4: Finding the final probability
The question asks for the probability that at least 2 items would be defective, but the answer choices are given in the form of 1 minus the probability. Therefore, we can calculate:
Final probability = 1 - P(at least 2)
Final probability = 1 - 0.1458
Final probability ≈ 0.8542

The closest answer choice to 0.8542 is option A, which is 1 - 0.94 × 1.4.
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