Probability that both are defective

[r0bHadz]

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?

 

[BvU]

No. The two is there because there are two ways to pick defective chips x and y from a set of 6: x first and then y or y first and then x.

 

[r0bHadz]

No. The two is there because there are two ways to pick defective chips x and y from a set of 6: x first and then y or y first and then x.

Hmm not sure what you mean mate? If you mean 2/6, that is the probability of your first draw being defective

 

[BvU]

Google 'combination'. There are $\left ( \matrix{6 \\ 2}\right )$ different combinations of two chips, of which one satisfies the requirement that both be defective.

$\left ( \matrix{6 \\ 2}\right ) = {6! \over 4! 2!} = {6\cdot 5\over 2}$

 

[r0bHadz]

Hmm i just thought of this using conditional probability and I think I might be on the right track now?

Event A =first pick is defective
P(A) = 2/6
Event B = second pick is defective
But B depends on A

P(B|A) = p(a intersect b)/p(a)

So since we know p(a) we can multiply p(b|a) by p(a) and get the intersect, which is what this problem is asking for

 

[RPinPA]

Your way of counting in your original solution is OK, but they're not independent.
Let D1 = the probability that the first chip you pull without replacement is defective.
Let D2 = the probability that the second chip you pull without replacement is defective.

You're thinking that you calculated P(D1 and D2) = P(D1) P(D2) but you didn't. You do indeed want P(D1 and D2) and you can calculate it from:
P(D2 | D1) = P(D1 and D2) / P(D1)
P(D1 and D2) = P(D1) P(D2 | D1)
The probability 1/5 that you used is the probability of drawing a defective chip given that the first one was a defective chip. Clearly whether or not the first chip was defective changes the probability of drawing a defective chip on the second draw, so they are clearly not independent. But as the equation above shows, multiplying those two probabilities was the right thing to do.

In this case, the conditional probability is easier to reason out than the unconditioned probability P(D2), but it's not always that way.

What would P(D2) be? That would be the fraction of all possible draws in which the second chip is defective, no matter what the first one was. That is, it's the probability of drawing "defective-defective" or "working-defective" in that order. You can evaluate that by conditioning on D1.

P(D2) = P(D2 | D1) P(D1) + P(D2 | ~D1) P(~D1)
= (1/5) (2/6) + (2/5) (4/6)

"Conditioning on [other event]" is a very powerful technique which often comes in handy.

Edit: And I see you posted while I was writing this, already realizing that you were working with conditional probability.

 

[Ray Vickson]

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 $P(D_1 \cap D_2) = (2/6) (1/5).$

 

[r0bHadz]

Great insight everyone! I got a lot of help from all of you.