A bin of nine parts contains seven good parts and two defective parts. To find the defective parts, we test them one-by-one, at random and without replacement. What is the probability that we are lucky and find both of the defective parts in the first two tests?
Question
A bin of nine parts contains seven good parts and two defective parts. To find the defective parts, we test them one-by-one, at random and without replacement. What is the probability that we are lucky and find both of the defective parts in the first two tests?
Solution
To solve this problem, we need to understand that it is a problem of combinations.
Step 1: Calculate the total number of ways to draw 2 parts from 9. This is a combination problem, which can be calculated using the formula for combinations: C(n, k) = n! / [k!(n-k)!], where n is the total number of items, k is the number of items to choose, and "!" denotes factorial.
So, the total number of ways to draw 2 parts from 9 is C(9, 2) = 9! / [2!(9-2)!] = 36.
Step 2: Calculate the number of ways to draw 2 defective parts from 2. This is also a combination problem, and can be calculated in the same way: C(2, 2) = 2! / [2!(2-2)!] = 1.
Step 3: The probability of drawing 2 defective parts in the first two tests is the number of ways to draw 2 defective parts divided by the total number of ways to draw 2 parts. So, the probability is 1 / 36 = 0.0278, or about 2.78%.
So, the probability that we are lucky and find both of the defective parts in the first two tests is about 2.78%.
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