Reply With QuoteTis a similar problem to one of the questions I asked if you want to take a look at my thread. They suggested a contingency table and it worked for me
Hi Guys
I'm new here so Hi to everyone.
Anyway I have done the rest of the questions easily but stuck on this one:
“Cascade Paint” mixes paint at its three separate plants, A, B and C. The unmarked paint cans are then sent to a central warehouse. Plant A supplies 42% of the paint and past records indicate that the paint is incorrectly mixed (defective) 1.5% of the time at this plant. Plant B supplies 38% of the paint with a defective rate of 1.8% and Plant C supplies 20% of the paint with a defective rate of 2.5%.
(a) If a can of paint is selected at random from the warehouse, what is the probability that it will be defective?
(b) Suppose that a can of paint selected at random from the warehouse is found to be defective. What is the probability that Plant A supplied it?
Please any help would really really be appreciated.
Cheers
Tis a similar problem to one of the questions I asked if you want to take a look at my thread. They suggested a contingency table and it worked for me
Hi guys again I believe I have figured it out but just need the formula to show the working out for the first part. So any help here would be welcomed.
The possibility of selecting a defective part is: 0.018
The second part I have done like this:
Let A = Plant A defective tins portion (.006)
Let D = Total defective tins (0.018)
P(A|D) = P(A and B) /P(D)
= 0.006/0.018
=0.35
Is this correct?
I think you have it, except for a typo in P(A and D)?
good job
jerry
Cheers man thanks, so instead of P(A and D) it should be P(A) * P(D) is that correct?Originally Posted by jerryb
sorry, you typed P(A and B) when i think you wanted P(A and D), since there was no B in your problem i presumed that to be a typo.
jerry
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