What is the probability a piece of clothing was made by person 1 if it is defective

Question

Person 1,2 and 3 produce the following proportions of clothes:

Person 1: 10%

Person 2: 30%

Person 3: 60%

The probability they make clothes that are defective are:

Persion 1: 4%

Person 2: 3%

Person 3: 2%

What is the probability a random piece of clothing was made by person 1 if it's defective

I have tried to use this formula but I'm not sure if my reasoning for intersections is correct:

$$ P(F_i|E) =\frac{P(E|F_i) \times P(F_i)}{P(E|F_1) \times P(F_1) + P(E|F_2) \times P(F_2)+ ... + P(E|F_n) \times P(F_n)}$$

So we solve for $P(\text{Person 1|Defective)}$

So, for example for the numerator we'd have $P(\text{Defective|Person 1}) \times P(\text{Person 1})$ $$ P(\text{Defective|Person 1}) = \frac{P(\text{Defective}\cap \text{Person 1})}{P(\text{Person 1})} = \frac{4\% \cap 10\%}{10\%} = \frac{4\%}{10%}$$

I simplified the intersection to this because i visualized person 1 as one big circle labelled with a 10% and the 4% being within that 10%

Plugging that value back into the formula in the numerator, I get:

$$ \frac{4\%}{10\%} \times 10\% = 4\%$$

Is this correct? If I am sure of this then I can finish the problem with more confidence.

Answer 1

There is no need to calculate $\mathsf P(E\mid F_1)$.   It is what you were given as the probability of a unit of clothes being defective when it is made by person 1.

You have been given : $\mathsf P(F_1)= 10\% \\ \mathsf P(F_2)=30\% \\ \mathsf P(F_3)=60\%\\\mathsf P(E\mid F_1)=4\%\\\mathsf P(E\mid F_2)= 3\% \\\mathsf P(E\mid F_3)=2\%$

Plug them into the formula to obtain: $$\mathsf P(F_1\mid E) = \dfrac{\mathsf P(E\mid F_1)~\mathsf P(F_1)}{\mathsf P(E\mid F_1)~\mathsf P(F_1)+\mathsf P(E\mid F_2)~\mathsf P(F_2)+\mathsf P(E\mid F_3)~\mathsf P(F_3)}$$

Answer 2

Let's consider a sample case. Persons 1, 2, and 3 make a total of 1000 shirts.

This means that:

• Person 1 makes 100 shirts.
• Person 2 makes 300 shirts.
• Person 3 makes 600 shirts.

Now, each of them make some defective shirts. Using the percentages you listed above:

• Person 1 makes 100*0.04 = 4 defective shirts.
• Person 2 makes 300*0.03 = 9 defective shirts.
• Person 3 makes 600*0.02 = 12 defective shirts.

So, there are 4 + 9 + 12 = 25 defective shirts in total.

Of these defective shirts, we're looking for the probability that Person 1 was the unlucky person who made the shirt we pulled.

Well, out of the 25 defective shirts, Person 1 made 4 of them, so 4/25 = 16%.

This is how I like to approach conditional probability problems - consider a fixed number x shirts were made, and bash the probabilities out.

The formula essentially does all the steps we just did in one fell swoop. (:

Let's also take a look at the formula just so you can get some intuition for what it's doing.

The left hand side is $P(F_i|E)$, which in this case is $P(Person 1|Defect)$.

The numerator of the right hand side is $P(E|F_i)*P(F_i)$, which in this case is $P(Defect|Person 1)*P(Person 1)$.

Well, we know $P(Defect|Person 1) = 0.04$ and we know $P(Person 1) = 0.1$.

Likewise, $P(Defect|Person 2) = 0.03$, $P(Person 2) = 0.3$.

Likewise, $P(Defect|Person 3) = 0.02$, $P(Person 2) = 0.6$.

Putting this all together, we get:

$P(Person 1|Defect) = (0.04*0.1)/(0.04*0.1 + 0.03*0.3 + 0.02*0.6) = 0.004/0.025 = 4/25$

Answer 3

I don't believe that is correct.

I will change the letters a little. Let $A_i$ be the event that person $i$ made the cloth, for $i = 1,2,3$, and let $D$, be the event that a cloth was defective. then we are interested in $$P(A_1|D) = \frac{P(A_1,D)}{P(D)}.$$ Regarding the numerator, we want to condition on $A_1$, $$P(A_1,D) = P(D|A_1)P(A_1) = .04(.10) = 0.004.$$

Below is my work $$$$$$$$$$$$$$$$ \begin{align*} P(P(A_1|D) &= \frac{P(A_1D)}{P(D)}\\ &=\frac{P(D|A_1)P(A_1)}{P(A_1D)+P(A_2D)+P(A_3D)}\\ &=\frac{P(D|A_1)P(A_1)}{P(D|A_1)P(A_1)+P(D|A_2)P(A_2)+P(D|A_3)P(A_3)}\tag 1\\ &=\frac{.04(.10)}{.04(.10)+.03(.3)+.02(.6)}\\ &=\frac{4}{25} \end{align*} where in $(1)$ I used the law of total probability in the denominator.