A box contains 15 pens of which 4 are defective. A random sample of 3 pens is taken without replacement and without regard to order. What is the probability of selecting at least one defective pen?
What you want in the probability of selecting 1, 2 or 3 defective pens.
P of 3 = 4/15 * 3/14 * 2/13
P of 2 = 4/15 * 3/14 * 11/13
P of 1 = 4/15 * 11/14 * 10/13
The answer is the sum of those three products.
0.2442
To calculate the probability of selecting at least one defective pen from a random sample of 3 pens, we need to calculate the probability of two scenarios: selecting exactly one defective pen, and selecting more than one defective pen.
Let's start with the scenario of selecting exactly one defective pen. There are 4 defective pens in the box, and we need to choose one of them. The total number of pens in the box is 15. So, the probability of selecting exactly one defective pen can be calculated as:
(Picking one defective pen) * (Picking two non-defective pens)
= (4/15) * (11/14) * (10/13)
Next, let's calculate the probability of selecting more than one defective pen. There are 4 defective pens in the box, and we need to choose two or three of them. The total number of pens in the box is 15. So, the probability of selecting more than one defective pen can be calculated as:
(Picking two defective pens) * (Picking one non-defective pen)
= (4/15) * (3/14) * (11/13)
Now, we can calculate the overall probability of selecting at least one defective pen by adding the probabilities of selecting exactly one defective pen and selecting more than one defective pen.
Probability of selecting at least one defective pen
= Probability of selecting exactly one defective pen + Probability of selecting more than one defective pen
= (4/15) * (11/14) * (10/13) + (4/15) * (3/14) * (11/13)
By simplifying this expression, we can find the final answer.