One company has $3$ locations where they produce calculators. In
total, the company produced $10000$ calculators. $2750$ of them were
produced in the first location, $4300$ in second location and $2950$
at the third location. From $2750$ produced calculators in first
location, $59$ are defective. In the second location, $85$ of $4300$
produced calculators are defective and in the third location $125$ of
$2950$. Determine the probability that you pick a defective calculator
from the $10000$ produced calculators.
I am not sure how this question can be solved. At first, I just concentrated on each location itself.
If we only look at location one, we have a probability of $\frac{59}{2750}$ to a defective calculator.
In location two we have $\frac{85}{4300}$ and in location three we have $\frac{125}{2950}$.
However, now we want the probability to pick one defective calculator from $10000$ produced calculators. I don't know if my logic is correct but I would simply sum up the single probabilities I just calculated, so we have
$$\frac{59}{2750}+\frac{85}{4300}+\frac{125}{2950} \approx 0.08359 \approx 8.36\text{%}$$
If I just look at the result, it does not seem very realistic. Should the answer not be lower?
Best Answer
There are $10000$ calculators, $59+85+125 = 269$ of which are defective. The probability of picking a defective calculator thus equals:
$$\frac{269}{10000} = 0.0269$$
If $A_i$ is the event in which a calculator is produced in factory $i$ and $D$ is the event in which a calculator is defective, this probability can also be calculated as:
$$P[D] = P[A_1] P[D | A_1] + P[A_2] P[D | A_2] + P[A_3] P[D | A_3] =$$
$$\frac{2750}{10000}\frac{59}{2750} + \frac{4300}{10000}\frac{85}{4300} + \frac{2950}{10000}\frac{125}{2950} = \frac{269}{10000} = 0.0269$$