# probability that there is at least one defective part

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.

## 3 Answers

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

• The probability that none is defective is still 0.94. So I have to: 0.94 * 0.94 * 0.94 * 0.94 = 0.7807, and this way I find the probability that I pick up no defective piece - If I understand this correctly.
– PMS
Jan 27, 2014 at 20:10
• That's right! Now use the same reasoning in b) to find $\mathbb{P}\{\text{No defective in 60 draws}\}$ which, together with the hint, gives you the answer.
– KOE
Jan 27, 2014 at 20:15
• Now I have to find at least one defective part. So: P(X=at least one defective) = 1 - P(non defective)^60 = 1 - 0.94^60 = 1 - 0.0224 = 0.97558, which is also the answer I had found. Can it be the right one?
– PMS
Jan 27, 2014 at 20:22
• Yes, that's right as well, well done!
– KOE
Jan 27, 2014 at 20:25
• Thank you so much! Now it's much easier to understand, you are so polite. Really thank you :)
– PMS
Jan 27, 2014 at 20:27

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.

• Thank you very much, it's a very clear explanation :) Now I understand
– PMS
Jan 27, 2014 at 20:49
• You are welcome. You basically knew how to do both (but found the "harder" variant easier). I just wanted to bring it down a little more to basics. Jan 27, 2014 at 20:51

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:

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