Bayes' Theorem/Examples/Arbitrary Example 1
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Example of Use of Bayes' Theorem
Let box $1$ contain $10$ good screws and $2$ unslotted screws.
Let box $2$ contain $8$ good screws and $4$ unslotted screws.
Let a box be selected at random.
Let a screw chosen from that box prove to be unslotted.
What is the probability that it came from box $2$?
Solution
Let $A$ be the event: An unslotted screw is selected.
Let $B_1$ be the event: Screw is selected from box $1$.
Let $B_2$ be the event: Screw is selected from box $2$.
We have that:
\(\ds \map \Pr {B_1} = \map \Pr {B_2}\) | \(=\) | \(\ds \dfrac 1 2\) | ||||||||||||
\(\ds \condprob A {B_1}\) | \(=\) | \(\ds \dfrac 1 6\) | ||||||||||||
\(\ds \condprob A {B_2}\) | \(=\) | \(\ds \dfrac 1 3\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \condprob {B_2} A\) | \(=\) | \(\ds \dfrac {\map \Pr {B_2} \condprob A {B_2} } {\map \Pr {B_1} \condprob A {B_1} + \map \Pr {B_2} \condprob A {B_2} }\) | Bayes' Theorem | ||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\frac 1 2 \times \frac 1 3} {\frac 1 2 \times \frac 1 6 + \frac 1 2 \times \frac 1 3}\) | plugging in the numbers | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 2 3\) | calculation |
Hence the probability is $\dfrac 2 3$ that the unslotted screw came from box $2$.
$\blacksquare$
Sources
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): Bayes' theorem
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Bayes' theorem