Question

If a machine is correctly set up, it produces $90\%$ acceptable items. If it is incorrectly set up, it produces only $40\%$ acceptable items. Past experience shows that $80\%$ of the set ups are correctly done. If after a certain set up, the machine produces $2$ acceptable items, find the probability that the machine is correctly set up.

Answer 1

Here we need to find the probability of the machine being correctly set up when the event of production of $2$ acceptable item has already happened. You can use the concept of conditional probability. Denote the given probabilities and find the conditional probability for getting two acceptable items when machine setting is correct. Now use the Bayes’ theorem for the required probability.

Complete step-by-step answer:
Let’s analyse the given information in the question. The chances or probability of a machine producing acceptable items when set up correctly is $90\% ,{\text{ i}}{\text{.e}}{\text{. }}\dfrac{{90}}{{100}} = 0.9$ . The chances or probability of a machine producing acceptable items when setups are incorrectly done is $40\% {\text{, i}}{\text{.e}}{\text{. }}\dfrac{{40}}{{100}} = 0.4$ . And the chances or probability to set up a device correctly is $80\% ,{\text{ i}}{\text{.e}}{\text{. }}\dfrac{{80}}{{100}} = 0.8$ .
With this information, we need to find the probability that the machine is correctly set up when the machine already produces $2$ acceptable items.
Before starting with the solution, you should understand the concept of conditional probability. The conditional probability of an event B is the probability that the event will occur given the knowledge that an event A has already occurred. This probability is written $P\left( {B|A} \right)$ , a notation for the probability of B given A. In the case where events A and B are independent (where event A does not affect the probability of event B), the conditional probability of event B given event A is simply the probability of event B, that is $P\left( B \right)$.
So, we let’s assume M to be the event of a machine producing $2$ acceptable items. The event of the machine being set up correctly be C and the event of the machine being set up incorrectly be N. Then, the probability of setting up the machine correctly is $P\left( C \right) = 0.8$ and the probability of setting up machine incorrectly is $P\left( N \right) = 1 - P\left( C \right) = 1 - 0.8 = 0.2$ .
And the conditional probability of getting two acceptable items when machine setting is correct will be $P\left( {M|C} \right) = 0.9 \times 0.9 = 0.81$ . Similarly, the probability of getting two acceptable items with the setting of machines being incorrect will be $P\left( {M|N} \right) = 0.4 \times 0.4 = 0.16$
Now we need to find $P\left( {C|M} \right)$ , i.e. the probability of the machine set up correctly given that $2$ items produced are acceptable.
For this, we can use Bayes’ theorem. Bayes' theorem is a formula that describes how to update the probabilities of hypotheses when given evidence. It follows simply from the axioms of conditional probability but can be used to powerfully reason about a wide range of problems involving belief updates.
Given a hypothesis $H$ and evidence $E$ , Bayes' theorem states that the relationship between the probability of the hypothesis before getting the evidence $P\left( H \right)$ and the probability of the hypothesis after getting the evidence $P\left( {H|E} \right)$ is:
$\Rightarrow P\left( {H|E} \right) = \dfrac{{P\left( {E|H} \right)}}{{P\left( E \right)}} \times P\left( H \right)$
Using Bayes’ theorem we can write $P\left( {C|M} \right)$ as:
$\Rightarrow P\left( {C|M} \right) = \dfrac{{P\left( {M|C} \right) \times P\left( C \right)}}{{P\left( {M|C} \right) \times P\left( C \right) + P\left( {M|N} \right) \times P\left( N \right)}}$
Now we can substitute the known values into the above expression:
$\Rightarrow P\left( {C|M} \right) = \dfrac{{P\left( {M|C} \right) \times P\left( C \right)}}{{P\left( {M|C} \right) \times P\left( C \right) + P\left( {M|N} \right) \times P\left( N \right)}} = \dfrac{{0.8 \times 0.9 \times 0.9}}{{0.8 \times 0.9 \times 0.9 + 0.2 \times 0.4 \times 0.4}}$
Solving the above expression, we get:
$\Rightarrow P\left( {C|M} \right) = \dfrac{{0.648}}{{0.648 + 0.032}} = \dfrac{{0.648}}{{0.68}} = \dfrac{{648}}{{680}} = 0.95$
So, we got the required probability as $0.95$

Note: The probability of an event E is denoted by $P\left( E \right)$ . Here when we calculated the conditional probability $P\left( {M|C} \right)$ and $P\left( {M|N} \right)$ we multiplied the probability twice since we required the answer for $2$ items and the given information was about just one acceptable item. Be careful while solving the Bayes’ theorem expression and use proper denotation to avoid complications.