Probability defective process

conditional probabilityprobabilityprobability distributionsprobability theory

Certain mechanical pieces undergo two independent processes by two distinct machines A and B. Stain A introduces defects in 5% of the processed pieces. a) Knowing that 7% of the final pieces are defective (ie that they have undergone both processes), calculate the value of the probability P (DB) that machine B introduces defects. b) If a randomly selected final piece is defective, calculate the probability that it has suffered defects from both machines. To carry out the procedure, use the following events: DA = {Machine A has introduced a defect} DB = {Machine B has introduced a defect} D = {The final piece is defective} {Hint: first express D using DA and DB}

I found that $ P(D_B) = 0,09 $ that’s because $ P(D) = P(D|D_A ) P(D_A) + P(D|D_B) P(D_B) $ but I think this is not correct because I also should have $ + P(D|AB) P(AB) $ but I don’t understand how to find this , to find after $P_B$

Best Answer

$P(D_A) = 0.05, P(D) = 0.07$ and we need to find $P(D_B)$.

There are $3$ ways for a product to be defective - defect by machine $A$, defect by machine $B$ or by both. There is only one way for the product to be non-defective - no defect introduced by either of the machines. So we can go about our calculation in two ways. I will use the latter.

$(1-P(D_A)) \times (1-P(D_B)) = 1 - P(D)$

i.e $0.95 \times (1-P(D_B)) = 0.93 \implies P(D_B) \approx 0.021$

Now to find the conditional probability please note that

$P(D_A \cap D_B) = P(D_A) \times P(D_B)$ and you already know $P(D)$.

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[Math] probability that there is at least one defective part

As asked for, here are some hints on how to solve these problems. If that gets you nowhere, let me know and I'll expand.

a) If you pick one piece out of a pile of pieces the machine made, the probability of picking one which is defective is $0.06$, so the probability of picking one which is not is $1 - 0.06 =0.94$. After picking one piece, you put it back (with replacement) and pick again; what're the probabilities for the second pick?

b) Note that the probability of at least one defective is the same as $1-\mathbb{P}\{\text{No defective}\}$

[Math] Conditional Probability: Distinction between $P(Fri|Defective)$ and $P(Defective|Fri)$

So far you have properly identified $P(D|M), \dots, P(D|F).$ But you need to look at Bayes' Theorem and figure out how to get the 'inverse' or 'posterior' probability $P(F|D).$ Here is an outline; you should match the steps with Bayes' Theorem.

Notice that $P(D \cap F),\, P(D|F)$ and $P(F|D)$ are all probabilities involving events $D$ and $F.$ But they refer to different populations: $P(D \cap F)$ refers to the entire weekly production of ball bearings; $P(D|F)$ refers to the population of ball bearings made on Friday; and $P(F|D)$ refers to the population of all defective ball bearings, asking what proportion of them were made on Friday.

$P(D \cap M) = P(M)P(D|M) = .2(.08) = .16.$ This uses what is sometimes called the 'general multiplication rule'.

$P(D) = P(D \cap M) + P(D \cap Tu) + \cdots + P(D \cap F) = 0.064.$ This uses what is sometimes called the 'law of total probability'.

Using R statisical software as a calculator:

p.d = sum(.2*c(.08, .04, .04, .04, .12));  p.d
## 0.064

$P(F|D) = P(D \cap F)/P(D) = .024/.064 = 0.375.$ This uses Bayes' Theorem.

While only 20% of all ball bearings are made on Friday, 37.5% of defective ball bearings were made on Friday.

Best Answer

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[Math] probability that there is at least one defective part

[Math] Conditional Probability: Distinction between $P(Fri|Defective)$ and $P(Defective|Fri)$

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