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I am trying to solve the problem 22 from Chapter 10 of "Chartrand, Gary & Zhang, Ping. Discrete Mathematics. Waveland Press, 2011." (pg. 374).

Three dice are tossed. What is the probability that 1 was obtained on two of the dice given that the sum of the numbers on the three dice is 7?

This same question was asked before here. So, I already know the answer. But I want to see it being solved with Bayes' Theorem explicitly.

Let,

$$ A: \text{one was obtained on two of the dice} $$ $$ B: \text{sum of the numbers on the three dice is 7} $$

Then, $$ P(B|A) = \frac{3}{6^3} $$ And, $$ P(A) = \frac{3\cdot 5}{6^3} $$ Finally, $$ P(B) = \frac{15}{6^3} $$ Because the ways we could get the sum of the numbers on the three dice to be seven is the coefficient of $x^7$ in $(x+x^2+x^3+x^4+x^5+x^6)^3$ which is 15.

Putting it all together, $$ P(A|B) = \frac{P(B|A)\cdot P(A)}{P(B)} = \frac{\frac{3}{6^3}\cdot \frac{3\cdot 5}{6^3}}{\frac{15}{6^3}} = \frac{3}{6^3}\cdot \frac{3\cdot 5}{6^3}\cdot\frac{6^3}{15} = \frac{3}{6^3} $$

And that is not the right answer. Note that I don't just want the right answer or a better way to get it, I want to know how Bayes' Theorem will work here.

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  • $\begingroup$ $\displaystyle \frac{\frac{3}{6^3}}{\frac{15}{6^3}} = \frac{3}{15}.$ $\endgroup$ Apr 19, 2023 at 1:13
  • $\begingroup$ Care to elaborate? Which of the two probabilities is one and why? $\endgroup$
    – scribe
    Apr 19, 2023 at 1:23
  • $\begingroup$ When you calculated $P(A)=\dfrac{3\cdot 5}{6^3}$, among the $3\cdot 5$ dice outcomes that satisfy $A$, how many dice outcomes satisfy $B$: "sum of the numbers on the three dice is $7$"? From this, I don't get your denominator $6^3$ of your $P(B\mid A)$. $\endgroup$
    – peterwhy
    Apr 19, 2023 at 1:26
  • $\begingroup$ Well, the sum of seven can only be achieved given two ones as $1,1,5$ and $1,5,1$ and $5,1,1$. This is $P(B|A) = 3/6^3$. $\endgroup$
    – scribe
    Apr 19, 2023 at 2:09
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    $\begingroup$ Your $3/6^3$ would be $P(B\cap A)$, and you got it correctly by listing all satisfying outcomes. But for $P(B\mid A)$, the denominator is not the full $6^3$, but $$P(B\mid A) = \frac{P(B\cap A)}{P(A)}$$ $\endgroup$
    – peterwhy
    Apr 19, 2023 at 2:25

1 Answer 1

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For a beginner to Bayes' Theorem, it is often useful to

  • think in terms of favorable sample space and applicable sample space

  • if you want to use probabilities, use what I call the "baby" Bayes' formula

Applied to this particular problem,

  • favorable sample space $=3$,
  • applicable sample space $=15$

and you can get the answer directly as $Pr = \dfrac3{15}$

And for using the "baby" Bayes' formula,

  • let $A$ = two out of three die faces are $1$,
  • let $B$ = sum of three die faces are $7$,

$P(A|B) = \dfrac{P(A \cap B)}{P(B)}= \dfrac3 {6^3}/\dfrac{15}{6^3} = \dfrac3{15}$

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