This problem can be solved using the Binomial Distribution. The binomial distribution model deals with finding the probability of success of an event which has only two possible outcomes in a series of experiments. In this case, we have a 'success' if a blade is defective.

(i) No defective blades in a packet:

The probability of a blade being defective is 0.002. Therefore, the probability of a blade not being defective is 1 - 0.002 = 0.998.

Since there are 10 blades in a packet, the probability of a packet having no defective blades is (0.998)^10 = 0.9801.

Therefore, the number of packets with no defective blades in a consignment of 50,000 packets is 0.9801 * 50,000 = 49,005 packets.

(ii) One defective blade in a packet:

The probability of exactly one blade being defective and the other nine not being defective in a packet is given by the binomial probability formula:

P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))

where n is the number of trials (10 in this case), k is the number of successes (1 in this case), p is the probability of success (0.002 in this case), and C(n, k) is the binomial coefficient.

So, P(X=1) = C(10, 1) * (0.002^1) * ((1-0.002)^(10-1)) = 10 * 0.002 * (0.998)^9 = 0.0179.

Therefore, the number of packets with one defective blade in a consignment of 50,000 packets is 0.0179 * 50,000 = 895 packets.

(iii) Two defective blades in a packet:

Similarly, the probability of exactly two blades being defective and the other eight not being defective in a packet is given by the binomial probability formula:

P(X=2) = C(10, 2) * (0.002^2) * ((1-0.002)^(10-2)) = 45 * 0.000004 * (0.998)^8 = 0.0003.

Therefore, the number of packets with two defective blades in a consignment of 50,000 packets is 0.0003 * 50,000 = 15 packets.

Question

This problem can be solved using the Binomial Distribution. The binomial distribution model deals with finding the probability of success of an event which has only two possible outcomes in a series of experiments. In this case, we have a 'success' if a blade is defective.

(i) No defective blades in a packet:

The probability of a blade being defective is 0.002. Therefore, the probability of a blade not being defective is 1 - 0.002 = 0.998.

Since there are 10 blades in a packet, the probability of a packet having no defective blades is (0.998)^10 = 0.9801.

Therefore, the number of packets with no defective blades in a consignment of 50,000 packets is 0.9801 * 50,000 = 49,005 packets.

(ii) One defective blade in a packet:

The probability of exactly one blade being defective and the other nine not being defective in a packet is given by the binomial probability formula:

P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))

where n is the number of trials (10 in this case), k is the number of successes (1 in this case), p is the probability of success (0.002 in this case), and C(n, k) is the binomial coefficient.

So, P(X=1) = C(10, 1) * (0.002^1) * ((1-0.002)^(10-1)) = 10 * 0.002 * (0.998)^9 = 0.0179.

Therefore, the number of packets with one defective blade in a consignment of 50,000 packets is 0.0179 * 50,000 = 895 packets.

(iii) Two defective blades in a packet:

Similarly, the probability of exactly two blades being defective and the other eight not being defective in a packet is given by the binomial probability formula:

P(X=2) = C(10, 2) * (0.002^2) * ((1-0.002)^(10-2)) = 45 * 0.000004 * (0.998)^8 = 0.0003.

Therefore, the number of packets with two defective blades in a consignment of 50,000 packets is 0.0003 * 50,000 = 15 packets.

Knowee AI · Accepted Answer