What is the chance of getting at least one defective item if 3 items a...
Solution:
To find the probability of getting at least one defective item, we need to find the probability of getting 1, 2, or 3 defective items.
Probability of getting 1 defective item out of 3 drawn:
The first defective item can be drawn in 2C1 ways (selecting 1 item out of 2 defective items), and the remaining 2 non-defective items can be drawn in 4C2 ways (selecting 2 items out of 4 non-defective items). Therefore, the probability of getting 1 defective item is:
P(1) = (2C1 x 4C2) / 6C3 = (2 x 6) / 20 = 0.6
Probability of getting 2 defective items out of 3 drawn:
The two defective items can be drawn in 2C2 ways (selecting 2 items out of 2 defective items), and the remaining 1 non-defective item can be drawn in 4C1 ways (selecting 1 item out of 4 non-defective items). Therefore, the probability of getting 2 defective items is:
P(2) = (2C2 x 4C1) / 6C3 = (1 x 4) / 20 = 0.2
Probability of getting 3 defective items out of 3 drawn:
The three defective items can be drawn in 2C3 ways (selecting 3 items out of 2 defective items, which is not possible), so the probability of getting 3 defective items is 0.
Therefore, the probability of getting at least one defective item is:
P(at least 1) = P(1) + P(2) + P(3) = 0.6 + 0.2 + 0 = 0.8
Hence, the correct answer is option C, 0.8.
What is the chance of getting at least one defective item if 3 items a...
Probability =1-4C3/6C3
1 -4/20
20-4/20
16/20
answer=0.80