Question

In a production line of a factory, each packet contains four items. Past record shows that 20% of the produced items are defective. A quality manager inspects each item in a packet and approves the packet for shipment if at most one item in the packet is found to be defective. Then the probability (round off to 2 decimal places) that out of the three randomly inspected packets at least two are approved for shipment equals ............

Hint:

For binomial probabilities, use the binomial probability mass function \( P(X = k) = \binom{n}{k} p^k (1 - p)^{n-k} \) to find the likelihood of different outcomes.

Answer 1

Correct Answer: 0.88

Step 1: Define the probability of a defective item.
Since 20% of the items are defective, the probability that an item is defective is \( P(\text{defective}) = 0.2 \), and the probability that an item is not defective is \( P(\text{not defective}) = 0.8 \).
Step 2: Define the binomial distribution.
Let \( X \) be the number of defective items in a packet. Since each packet contains 4 items, \( X \) follows a binomial distribution with parameters \( n = 4 \) and \( p = 0.2 \), i.e., \( X \sim \text{Binomial}(4, 0.2) \).
The probability that at most one item is defective (i.e., \( X \leq 1 \)) in a packet is: \[ P(X \leq 1) = P(X = 0) + P(X = 1). \] Using the binomial probability formula: \[ P(X = k) = \binom{4}{k} p^k (1 - p)^{4-k}. \] For \( X = 0 \): \[ P(X = 0) = \binom{4}{0} (0.2)^0 (0.8)^4 = (0.8)^4 = 0.4096. \] For \( X = 1 \): \[ P(X = 1) = \binom{4}{1} (0.2)^1 (0.8)^3 = 4 \times 0.2 \times 0.512 = 0.4096. \] Thus, \[ P(X \leq 1) = 0.4096 + 0.4096 = 0.8192. \] Step 3: Find the probability for three packets.
Let \( Y \) be the number of approved packets. Since the packets are inspected independently, \( Y \) follows a binomial distribution with parameters \( n = 3 \) and \( p = 0.8192 \), i.e., \( Y \sim \text{Binomial}(3, 0.8192) \).
We want to find the probability that at least two packets are approved: \[ P(Y \geq 2) = P(Y = 2) + P(Y = 3). \] Using the binomial probability formula again: \[ P(Y = k) = \binom{3}{k} (0.8192)^k (1 - 0.8192)^{3-k}. \] For \( Y = 2 \): \[ P(Y = 2) = \binom{3}{2} (0.8192)^2 (0.1808)^1 = 3 \times 0.6717 \times 0.1808 = 0.3636. \] For \( Y = 3 \): \[ P(Y = 3) = \binom{3}{3} (0.8192)^3 (0.1808)^0 = 1 \times 0.5491 = 0.5491. \] Thus, \[ P(Y \geq 2) = 0.3636 + 0.5491 = 0.9127. \]
Final Answer: \[ \boxed{0.91}. \]