A lumber company has just taken delivery on a lot of 10,000 2 times 4 boards. Su

Question

A lumber company has just taken delivery on a lot of 10,000 2 times 4 boards. Suppose that 20% of these boards are actually too green to be used in first-quality construction. Two boards are selected at random, one after the other. Let A = {the first board is green} and B = {the second board is green}. Compute P(A), P(B) and P(AB). Are A and B independent? Now consider two independent events A and B with P(A) = P(B) =.2, what is P(AB)? How much difference is there between this answer and P(AB) in part (a)? For purposes of calculating P(AB), can we assume that A and B of part (a) are independent to obtain essentially the correct probability? Now Suppose the lot consists of ten boards, of which two are green. Does the assumption of independence now yield approximately the correct answer for P(AB)? What is the critical difference between the situation here and that of part (a)? When do you think that an independence assumption would be valid in obtaining an approximately correct answer to P{AB)?

Explanation / Answer

(a)

Let

A = {the first board is green}

B = {the second board is green}

Using the definition of probability, we may calculate the probability that the first board

Selected is green as

P(A) = Number of green boards / Total number of boards =2000/10000 =0.2

P(A) = 0.2

Because there is no replacement in this experiment, the probability that the second board is Green depends on whether or not the first board selected was green. Intuitively, we may already infer that A and B are not independent! Let’s define A1 as the complement of A.

A¢ = {the first board is not green}

According to the definition of complementary events,

P(A1) = 1 – P(A) = 1 – 0.2 = 0.8

The definition of probability provides us with information pertaining to the probability that

the second board selected is green. The two possible ways that event B may occur are

P(BA) = N(BA) N-1

             =Total number of green boards after A total number of boards after one selection

            =1999/9999

            =0.19992

And

P(BA1) = N(BA1)N-1

            =Total number of green boards after A1 total number of boards after one selection

            =20009999

            =0.20002

The probability that the second board selected is green may be found by applying the total

law of probability and using everything we’ve defined so far

P(B) = P(BA) * P(A) +   P(BA1) * P(A1)

         =19999999 * 0.2 + 20009999 *0.8

          =0.2

Now, we can finally calculate the intersection of events A and B. Actually, we

just did that in the solution for P(B)! The probability that the first two boards selected are

green may be found using the multiplication rule.

        P(A intersection B) = P(B | A) ×B =19999999 * 0.2 =0.039984

It was previously mentioned that the events A and B are not independent. We may verify

this by establishing the following inequality, which is a violation of the definition of

independence.

                P(A intersection B) = 0.039984 not equal to 0.04 = P(A) ×P(B)

(b)

Assuming the independence of A and B, we may calculate the intersection of A and B as
P(A intersection B) = P(A)×P(B) = 0.2 × 0.2 = 0.04
There is little difference between this answer and that computed in part (a). The difference
is 0.000016. Because this number is so small, it is quite reasonable to assume that A and B
are independent for purposes of simplifying future calculations.

(c)

Now the total number of boards, N = 10. Without assuming independence, we may
calculate the intersection of A and B in the same fashion as part (a). Note that the
probability that A occurs is still equal to 0.2.
P(A intersection B) = P(B | A) ×P(A) = 1/9 * 0.2 =0.2222
Now, making the assumption of independence would be a big mistake! The
difference between the solutions in parts (c) and (b) is 0.01778. Let’s figure out why the
approximation became a bad one once the population size decreased…

By comparing the experiments described in parts (a) and (c), we see that the validity of the
assumption of independence improves as the population size increases relative to the
sample size.
. It would make our lives easier if we were able to assume that an
experiment has binomial properties whenever possible. A rule of thumb exists for the
appropriate application of such an assumption.
The rule states that, in an experiment such as the one described in this problem, as long as
the sampling size is less than 5% of the sample population, the assumption of
independence is appropriate. So there’s our answer!
When given an experiment for which the assumption of independence would be useful, we
may assume independence as long as the sampling size is much smaller (i.e. <5%) of the
sample population.

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