What is the probability that both products are out of stock?

[BrownianMan]

I'm unsure about these Qs, but I attempted it & posted my answers. If any could verify if I'm on the right track, i'd appreciate it!

A company has just placed an order with a supplier for two different products. Let A1 be the event that the first product is out of stock and let A2 be the event that the second product is out of stock. Suppose that P(A1) = 0.3, P(A2) = 0.2, and the probability that at least one product is out of stock is 0.4.

(a) What is the probability that both products are out of stock?
(b) Are A1 and A2 independent events? Explain.
(c) Given that the first product is in stock, what is the probability that
the second is also?

My answers

(a) (0.3)(0.2) = 0.06
(b) Yes, they are independent
(c) P(A2 | A1) = P(A2) = 0.2

A construction firm has bid on two different contracts. Let A1 be the event that the bid on the first contract is successful, and define A2 analogously for the second contract. Suppose that P(A1) = 0.4, P(A2) = 0.3, and that A1 and A2 are independent events.

(a) Calculate the probability that both bids are successful.
(b) Calculate the probability that neither bid is successful.
(c) What is the probability that the firm is successful in at least one of the
two bids?

My answers:

(a) (0.4)(0.3) = 0.12
(b) (1-0.4)(1-0.3) = 0.42
(c) 0.4+0.3-0.12 = 0.58

 

[sylas]

Second part looks ok. But the first...

(a) (0.3)(0.2) = 0.06
(b) Yes, they are independent
(c) P(A2 | A1) = P(A2) = 0.2

Your answer to (a) assumes independence.
Your answer to (b) gives no reason.
Your answer to (c) also assumes independence.

How do you know they are independent?

 

[BrownianMan]

The events are independent because P(A1) is the same, regardless of whether or not A2 occurs, and vice versa.

 

[sylas]

The events are independent because P(A1) is the same, regardless of whether or not A2 occurs, and vice versa.

P(A1) is 0.3; that is not the same as P(A1 given A2), or P(A1 given -A2). In fact, they are all different values given the numbers provided.

 

[BrownianMan]

So they are dependent?

I don't understand why they would be, given that they are two different products. Could you explain it, please?

 

[sylas]

So they are dependent?

I don't understand why they would be, given that they are two different products. Could you explain it, please?

Independence of events means that the probability of one is unaffected by the probability of the other.

In this case, perhaps there is a certain probability that a better company is bidding as well. If they are bidding, both your bids are likely to fail. If they are not, both are likely to succeed. So the probability that the second bid will fail is greater, given that the first bid fails.

Or perhaps there is another smaller company bidding, and they are only going to submit one bid. In this case, the probability that the second bid will fail is smaller, given that the first bid fails.

There are all kinds of ways events can be dependent on each other, in this sense.

What formulae do you know that could be applied here?

Cheers -- sylas

 

[BrownianMan]

Independence of events means that the probability of one is unaffected by the probability of the other.

In this case, perhaps there is a certain probability that a better company is bidding as well. If they are bidding, both your bids are likely to fail. If they are not, both are likely to succeed. So the probability that the second bid will fail is greater, given that the first bid fails.

Or perhaps there is another smaller company bidding, and they are only going to submit one bid. In this case, the probability that the second bid will fail is smaller, given that the first bid fails.

There are all kinds of ways events can be dependent on each other, in this sense.

What formulae do you know that could be applied here?

Cheers -- sylas

You said the second question was ok, and that there was only a problem with the first one. There's no bidding in the first question - just a company placing an order with a supplier for two different products.

I guess I should have made it clear that they were two different questions. Sorry.

 

[sylas]

You said the second question was ok, and that there was only a problem with the first one. There's no bidding in the first question - just a company placing an order with a supplier for two different products.

I guess I should have made it clear that they were two different questions. Sorry.

No problem. It's my fault -- I knew they were two different problems, and just picked my example poorly.

Suppose the reason that something is out of stock is that a truck breaks down. That means the probability one is out of stock implies a greater probability the other is out of stock. Suppose that the reason one is out of stock is a successful advertising campaign which means one is selling more than the other. That would imply a smaller probability the other is out of stock.

The point is that there's no reason here to assume independence.

Have got any formulae you can use that involve multiple events?

Here's an example.

Pr(x or y) = Pr(x) + Pr(y) - Pr(x and y)

Any more?

Cheers -- sylas

 

[BrownianMan]

OK, so for (a), since we know that P(A1 or A2) = P(A1) + P(A2) - P(A1 and A2), and we know that P(A1 or A2) = 0.4 (probability that at least one is out of stock), then

0.4 = (0.3) + (0.2) - P(A1 and A2)
P(A1 and A2) = (0.3) + (0.2) - 0.4
P(A1 and A2) = 0.1

Then for (c), P(A2 | A1) = P(A1 and A2) / P(A1) = 0.1 / 0.3 = 0.33

Is that right?

 

[sylas]

OK, so for (a), since we know that P(A1 or A2) = P(A1) + P(A2) - P(A1 and A2), and we know that P(A1 or A2) = 0.4 (probability that at least one is out of stock), then

0.4 = (0.3) + (0.2) - P(A1 and A2)
P(A1 and A2) = (0.3) + (0.2) - 0.4
P(A1 and A2) = 0.1

Then for (c), P(A2 | A1) = P(A1 and A2) / P(A1) = 0.1 / 0.3 = 0.33

Is that right?

Yes. Well done.

 

[BrownianMan]

Thank you so much!

 

[vela]

I think you calculated the wrong probability for part (c). The problem asked for the probability the item 2 is in stock given item 1 is also in stock, but you found the probability item 2 is out of stock given item 1 is also out of stock.

 

[sylas]

I think you calculated the wrong probability for part (c). The problem asked for the probability the item 2 is in stock given item 1 is also in stock, but you found the probability item 2 is out of stock given item 1 is also out of stock.

Quite right. I looked too quickly. Sorry!

 

[BrownianMan]

So would it be 0.67 then?

 

[sylas]

So would it be 0.67 then?

No. Back the equations, I'm afraid.

 

[BrownianMan]

I have a question about this formula: Pr(x or y) = Pr(x) + Pr(y) - Pr(x and y)

Doesn't it only apply to events that are independent? If so, then it shouldn't apply to the first question, right?

 

[sylas]

I have a question about this formula: Pr(x or y) = Pr(x) + Pr(y) - Pr(x and y)

Doesn't it only apply to events that are independent? If so, then it shouldn't apply to the first question, right?

That one applies everywhere. How would you prove it?

One method would be to express everything in terms of the following four cases.
Pr(x and y)
Pr x and not y)
Pr(not x and y)
Pr(not x and not y)

 

[BrownianMan]

Ok, thanks!

I'm going to try (c) again tomorrow.

 

[BrownianMan]

Ok, this is what I have for (c) now:

P(not A1 and not A2) = 1 - P(A1 or A2) = 1 - 0.4 = 0.6

P(not A2 | not A1) = P(not A1 and not A2) / P(not A1) = 0.6/0.7 = 0.86

 

[sylas]

Ok, this is what I have for (c) now:

P(not A1 and not A2) = 1 - P(A1 or A2) = 1 - 0.4 = 0.6

P(not A2 | not A1) = P(not A1 and not A2) / P(not A1) = 0.6/0.7 = 0.86

I think that's right. Speaking carefully after I screwed up last time...

 

[simonW]

I have question. follow the question. they say "the probability that at least one product
is out of stock is .4". mean the probability that both product are in stock is 1-0.4=0.6.
if A1 and A2 are independent. the probability that both product are in stock should satisfy
0.6=P(complement of A1)*P(complement of A2). but (1-0.3)*(1-0.2) =0.56 !=0.6.

so that my question.

my thought is right?

 

[vela]

Yes. The contradiction indicates that your assumption that A1 and A2 are independent is wrong.