What is the probability that on a given day there shelia will find defects

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I study economics and I am trying to solve this problem - without success, unfortunately.

A machine has a defect rate of 6%.

a) Chosen at random 4 pieces (with replacement) from the production flow, compute the probability that none is defective.
b) In the event that extractions are performed 60, compute the probability that there is at least one defective part.

I have no idea how to answer the first question. What logical processes can I use?

Regarding the second question, I thought of using the binomial random variable with $n = 60$ and $p = 0.06$. Then: $$\begin{multline} P (X \geq 1) = 1 - P (X < 1) = 1 - P (X = 0) = \\ = 1 - \binom{60}{0} \cdot (0.06)^ 0 \cdot (0.94)^{60} = 1 - 0.02446 = 0.97558.\end{multline}$$

Can somebody please explain me how to solve this problem? Thank you very much.

user642796

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asked Jan 27, 2014 at 19:53

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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 $1-\mathbb{P}\{\text{No defective}\}$

answered Jan 27, 2014 at 20:03

KOEKOE

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The probability that a piece is defective is $0.06$. Thus the probability it is not defective is $1-0.06$, that is, $0.94$.

The probability that all four items are non-defective is therefore $(0.94)^4$. For let $G_1$ be the event the first item is good, $G_2$ the event the second item is good, and so on up to $G_4$. Each of these events has probability $0.94$. The $G_i$ are independent. So the probability they all occur (that is, all four items are good) is the product of the individual probabilities, that is, $(0.94)(0.94)(0.94)(0.94)$.

The second problem uses the same ideas. We interpret "extractions are performed $60$" to mean we test $60$ items. The probability none is defective is $(0.94)^{60}$, and therefore the probability at least one is defective is $1-(0.94)^{60}$.

Remark: Your analysis and calculation for the second problem were correct. The same kind of analysis, but simpler, settles the first question. Let $Y$ be the number of defective in $4$ trials. Then $Y$ has binomial distribution, $p=0.06$, $n=4$. We want the probability that $Y=4$. This is $$\binom{3}{0}(0.06)^0(0.94)^4.$$ That gives the same answer as our previous calculation.

answered Jan 27, 2014 at 19:53

André NicolasAndré Nicolas

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If a piece has a $6\%$ defect rate then the probability of not getting any defects with $4$ pieces (with replacement) is $(1-0.06)^4 \approx 78.07\%.$

The probability of getting at least one defective piece is one minus the probability of getting no defective pieces:

$$P_{>0} = 1 - P_0 = 1 - 0.94^{60} \approx 2.44\%.$$

answered Jan 27, 2014 at 20:02

JohnJohn

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