Question

Three dice have the probabilities of throwing a "five" as p, q and r respectively. One of the dice is chosen at random (each is equally likely to be chosen) and thrown and a "five" appeare
D. What is the probability that the die chosen was the first one?

• A. $\frac{r}{p+q+r}$
• B. $\frac{p}{p+q+r}$
• C. $\frac{q}{p+q+r}$
• D. $\frac{p+q+r}{3}$

Hint:

When the prior probabilities are all equal (like 1/3 here), the Bayes' probability for any branch is just the likelihood of that branch divided by the sum of all likelihoods. This "equal priors" shortcut is very common in competitive exams.

Answer 1

The Correct Option is B

This is a classic inverse probability problem that requires the use of Bayes' Theorem.

Step 1: \color{redDefine the Events and Prior Probabilities
Let $D_1, D_2, D_3$ be the events of choosing the first, second, and third die respectively.
Since each is equally likely to be chosen: $P(D_1) = P(D_2) = P(D_3) = \frac{1}{3}$.
Let $A$ be the event that a "five" is thrown.

Step 2: \color{redIdentify the Likelihoods
The probability of throwing a five given a specific die is:
$P(A|D_1) = p$
$P(A|D_2) = q$
$P(A|D_3) = r$

Step 3: \color{redApply Bayes' Theorem Formula
We want to find $P(D_1|A)$:
$P(D_1|A) = \frac{P(A|D_1)P(D_1)}{P(A|D_1)P(D_1) + P(A|D_2)P(D_2) + P(A|D_3)P(D_3)}$.

Step 4: \color{redSubstitute and Simplify
$P(D_1|A) = \frac{p \cdot \frac{1}{3}}{p \cdot \frac{1}{3} + q \cdot \frac{1}{3} + r \cdot \frac{1}{3}}$.
Since $\frac{1}{3}$ is common to all terms in the numerator and denominator, it cancels out:
$P(D_1|A) = \frac{p}{p + q + r}$.
The probability is $\frac{p}{p+q+r}$, which is Option (2).