# Help applying the Bayes's Theorem.

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.

• $\displaystyle \frac{\frac{3}{6^3}}{\frac{15}{6^3}} = \frac{3}{15}.$ Apr 19, 2023 at 1:13
• Care to elaborate? Which of the two probabilities is one and why? Apr 19, 2023 at 1:23
• 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 $P(B\mid A)$. Apr 19, 2023 at 1:26
• 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 $P(B|A) = 3/6^3$. Apr 19, 2023 at 2:09
• 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 $$P(B\mid A) = \frac{P(B\cap A)}{P(A)}$$ Apr 19, 2023 at 2:25

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 $$7$$,

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