A shipment of 50 parts contains 12 defective parts. Suppose 3 parts are selected at random, without replacement, from the shipment. What is the probability that all 3 parts are defective?
If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
12/50 * (12-1)/(50-1) * (12-2)/(50-2) = ?
4.845885593028\times10^{18}
4.845885593028\times10^{18} sorry its by mistake.
thanks for your cooperation
To find the probability that all three parts are defective, we need to use the concept of conditional probability.
First, let's find the probability of selecting the first defective part. Since there are 12 defective parts out of 50 in total, the probability of selecting a defective part as the first pick is 12/50.
Next, since we are selecting without replacement, after the first defective part is picked, there are now 11 defective parts out of the remaining 49 parts. So the probability of selecting a defective part as the second pick is 11/49.
Finally, after the first and second defective parts are picked, there are 10 defective parts out of the remaining 48 parts. Therefore, the probability of selecting a defective part as the third pick is 10/48.
To find the probability that all three parts are defective, we multiply these probabilities together since they are independent events:
(12/50) * (11/49) * (10/48) ≈ 0.0081743869
Therefore, the probability that all three parts selected are defective is approximately 0.0082, or 0.82%.