r0bHadz said:
Homework Statement
Out of six computer chips, two are defective. If two chips are randomly chosen for testing
(without replacement), compute the probability that both of them are defective. List all
the outcomes in the sample space.
Homework Equations
The Attempt at a Solution
I want to know:
P(both draws are defective)
= P(draw one is defective and draw 2 is defective)
= P(draw one is defective ∩ draw 2 is defective)
I'm confused. intuitively 2/6 * 1/5 makes sense to me and this is indeed the answer. But I am trying to understand the subject in terms of definitions and formulas, and I end up getting lost..
I have "events e1,e2,...en are independent if they occur independently of each other, occurence of one events does not effect the probability of another"
well (1/5)(2/6) is the probability of a intersect of independent events. But I don't see how they could possibly be independent. If we said "with replacement" I could see the case being made, but the question says without replacement, so the probability of getting a defective chip on the first draw effects the probability of getting a defective chip on the second draw.
I have "events A and B are disjoint if A∩B= ∅ " and under my current definition of the outcomes for this set, they are not disjoint either, so they cannot be mutually exclusive. That leaves me with no formulas I can work with.
Maybe changing the outcomes from: {a defective chip is drawn, a non defective chip is drawn}
to
{first chip is defective and second is defective, first chip is defective and second is not defective, first chip is not defective and second chip is defective, first chip is not defective and second chip is not defective}
would make more sense?
There are numerous ways to approach such problems.
(1) Direct sample-space approach.
Suppose the chips are numbered 1-6, and chips 1,2 are the defectives. We can think of the sample space as being the set of all permutations of the numbers 1-6, but we pick only the first two. (Of course, that will give numerous "duplicate" points when we look only at positions 1 and 2, but f you think about it you will see that no double-counting is involved, provided that we count everything properly, as is done below. The total number of permutations is 6!. The number of permutations in which chip 1 is first and 2 is second is 4!. The number of permutations in which chip 2 is first and 1 is second is 4!. Therefore, the number of permutations in which 1 and 2 occur in the first two positions is 2*4!. Therefore, the probability P(choose both defectives) = 2*4!/6! = (2*1)/(6*5) = (2/6)*(1/5).
(2) Using "restricted" permutations.
Again,let the chips be numbered 1-6, with chips 1 and 2 defective. Let ##{}_nP_m ## denote the number of permutations of ##n## things taken ##m## at a time; that is, $${}_nP_m = n (n-1) (n-2) \cdots (n-m+1) = \frac{n!}{(n-m)!}$$ For the case of ##n=6## and ##m=2##, ##{}_2P_6## is the number of possible
ordered pairs ##(i,j), \: i \neq j## taken from ##\{1,2,3,4,5,6 \}.## The set of all such ordered pairs is the sample space. Exactly two points in the sample sample space correspond to the event of interest; that is,
$$\{\text{both defective} \} = \{(1,2), (2,1) \}$$ so
$$P(\text{both defective}) = \frac{2}{{}_2P_6} = \frac{2 \cdot 4!}{6!} = \frac{2}{6} \frac{1}{5}.$$
(3) Conditional probability methods. Let ##D_ 1= \{ \text{1st drawn chip is defective} \}## and ##D_2 = \{ \text{2nd drawn chip is defective} \}.## The answer we want is $$P(D_1 \cap D_2) = P(D_1) \cdot P(D_2 |D_2).$$ We have ##P(D_1) = 2/6,## because on the first draw, two of the six possibilities are defective. Now, goven that ##D_1## has occured, there remain 5 chips among which exactly one is defective, so ##P(D_2|D_1) = 1/5.## Altogether, we have ##P(D_1 \cap D_2) = (2/6) (1/5).##