this is my question, can someone help me with it .

Question

Q‒2. A box contains 24 transistors, 4 of which are defective. If 4 are sold at random, find the following probabilities.

a) Exactly 2 are defectives.

b) None is defective.  

c) All are defective.

d) At least 1 is defective.

Answer 1

Hello Peter

Your problem fits into the following general scenario. We have a finite set of objects, divided into 2 types, say "type A" and "type B". The "experiment" consists of choosing a sample from the given set (without replacement.) We then determine how many elements of the sample fall into one of the two categories (say type A). Probabilities in this type of experiment are given by the hypergeometric distribution.

Now let us relate the hypergeometric distribution to your specific problem.

Let N = number of transistors = 24.

Let A = number of defectives = 4. (Then the number of nondefectives is N - A = 20).

Let n = the sample size = number of transistors selected(sold) = 4.

Let x = number of defectives in the chosen sample.

Then the probability of getting x defectives in the given sample is given by the hypergeometric distribution.

P(x) = (_AC_x)(_N-AC_n-x)/(_NC_n)

In the above formula, the _AC_x represents the number of combinations of x objects chosen from a set of size A. You might have a combination key on your calculator. If not, _AC_x is given by

_AC_x = A!/[x!(A - x)!],

where ! is the factorial symbol.

Part a) Use the hypergeometric distribution formula, with N = 24, A = 4, n = 4, and x = 2.

P(x) = (₄C₂)(_{(24 - 4)}C_(4-2))/(₂₄C₄) = (₄C₂)(₂₀C₂)/(₂₄C₄)

P(2) = (6*190)/10626

P(2) = 1140/10626

P(2) = 190/1771 ≅ .1073

Part b) Use the same formula with x = 0. The result will be P(0) ≅ .456

Part c) Use x = 4.

Part d) "At least 1 is defective" means X ≥ 1. The complementary event for "at least one" is "None". That is, the complement of "X ≥ 1" is X = 0 (if X is a discrete random variable.) Therefore, the easiest way to get the result is as follows:

P(At least one defective) = 1 - P(None are defective)

P(At least one defective) = 1 - P(0)

P(At least one defective) = 1 - .456 [Using the result from part b.]

P(At least one defective) ≅ .544

Hope that helps! If you have any questions or need to see any additional details, please let me know.

William