Consider it as there is a box which contains 10 items, out of which 3 are defective and 7 are nondefective items. Out of these 10 items, we can choose 4 items in following ways: 10 C 4 = (10! / 4! 6!) =
210 ways.
X -> A random variable that denotes the number of defective items in a sample.
So X can take following values:
X = 0 (When no item is defective) :
This means that all the items that we will choose will be from the non-defective items. This can be done
in 7 C 4 ways = (7! / 4! 3!) = 35 ways.
So (x = 0) = 35/210 = 1/6 .
X = 1(When 1 item of the selected ones is defective and the other 3 are non-defective) :
This means that we need to choose 3 of the 4 items from non-defective items and the remaining 1 item
from defective items. This can be done in 7 C 3 * 3 C 1 = (7!/4! 3!) * (3!/2! 1!) = 35 * 3 ways
So (X=1) gives (35*3)/(210) = 1/2
llly for (X = 2): Probability = 3/10
and for (X = 3): Probability = 1/30.
A box contains 12 items of which 3 are defective. A sample of 3 items is selected from the box. What is the probability distribution of the number of
defective items?
Since you said “A sample of 3 items is selected from the box” I will infer that the 3 items in that sample are
drawn without replacement.
So, without replacement the answer is the Hypergeometric Distribution, not the Binomial Distribution :
https://www.probabilitycourse.com/chapter4/4_1_4_solved4_1.php