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Whilst studying for an upcoming exam on probability, (binomial, continuous, and normal distributions) i'm stuck on the following problem. For reference, the problem is from Chapter 19 in Cambridge Mathematical Methods, Unit 3&4


A factory has two machines that produce widgets. The time taken, $X$ seconds, to produce a widget using machine A is normally distributed, with a mean of $10$ seconds and a standard deviation of $2$ seconds. The time taken, $Y$ seconds, to produce a widget using machine B has probability density function given by

$$f(y) = \begin{cases}\frac{1}{8}(y-8) & 8 \lt y \lt 12 \\ 0 & otherwise \\ \end{cases}$$

Suppose that $60\%$ of the widgets manufactured at the factory are produced by machine A, and $40\%$ by machine B. If a widget selected at random is known to have been produced in less than $10$ seconds, what is the probability that it was produced by machine A?


Because Machine A is normally distributed, the probability that it is able to produce a widget in less then $10$ seconds is $1/2$. For machine B, i've integrated the PDF from $8$ to $10$ to find that the probability that it is able to produce a part in less than $10$ seconds is $1/4$.

I'm struggling with the second part of the question, could someone please provide an answer for how to do so.

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1 Answer 1

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It is just a conditional probability problem.

For A:

$$\mathbb{P}[X<10]=\Phi(0)=0.5=p_1$$

For B: use the given density and find $p_2=0.25$

The requested probability is

$$\frac{0.6p_1}{0.6p_1+0.4p_2}=0.75$$

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