Animated Solution for Mathematics - Probability: Comprehension Passage

Welcome, future engineer. Today, we are not just solving a probability problem; we are stepping into the shoes of a detective.

In probability, we often deal with 'forward' problems—where we know the cause and calculate the effect. But today, we are going to do something more thrilling: we are going to work backward from an effect to find its most likely cause.

This is the essence of Bayes' Theorem, and it is a tool that will serve you for the rest of your career.

The Setup

Imagine two urns sitting on your desk. Urn $U_1$ is rich, holding $3$ white and $2$ red balls. Urn $U_2$ is humble, holding only $1$ white ball.

We are about to perform an experiment, but the rules of this experiment are dictated by the flip of a fair coin. This coin toss is our 'branching point.'

If it lands on Heads, we transfer $1$ ball. If it lands on Tails, we transfer $2$. This is the foundation of our entire calculation.

The Head Path

Let us explore the universe where the coin shows Heads. The probability of this is $P(H)=\frac{1}{2}$.

If we are in this universe, we draw $1$ ball from $U_1$. The probability of drawing a white ball is $\frac{3}{5}$, and a red ball is $\frac{2}{5}$.

Now, look at $U_2$. If we transferred a white ball, $U_2$ now contains $2$ white balls and $0$ red balls. The probability of drawing a white ball from $U_2$ is now $1$.

If we transferred a red ball, $U_2$ contains $1$ white and $1$ red, making the probability $\frac{1}{2}$. Combining these, the conditional probability $P(W|H)$ is:

The Tail Path

Now, let us step into the second universe: the Tail path. Here, we transfer $2$ balls.

We must use combinations. The total ways to pick $2$ balls from $5$ is $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$.

The probability of picking $2$ white balls is $\frac{\left(\genfrac{}{}{0ex}{}{3}{2}\right)}{10}=\frac{3}{10}$. The probability of picking $2$ red balls is $\frac{\left(\genfrac{}{}{0ex}{}{2}{2}\right)}{10}=\frac{1}{10}$. The probability of picking $1$ white and $1$ red is $\frac{\left(\genfrac{}{}{0ex}{}{3}{1}\right)\cdot \left(\genfrac{}{}{0ex}{}{2}{1}\right)}{10}=\frac{6}{10}$.

Now, we check $U_2$ again. If $2$ white balls were transferred, $U_2$ has $3$ white balls ($P(W)=1$). If $2$ red balls were transferred, $U_2$ has $1$ white and $2$ red ($P(W)=\frac{1}{3}$). If $1$ of each was transferred, $U_2$ has $2$ white and $1$ red ($P(W)=\frac{2}{3}$).

Summing these gives us $P(W|T)$:

The Law of Total Probability

We have two universes. To find the total probability of drawing a white ball, we merge them using the Law of Total Probability:

Substituting our values, we get:

This is the probability that, regardless of the coin toss, you will draw a white ball from $U_2$.

The Detective Work (Bayes' Theorem)

Finally, we are told the ball drawn is white. We want to know: was it a Head? This is $P(H|W)$.

Bayes' Theorem tells us:

We have all the pieces. The numerator is $\frac{1}{2}\cdot \frac{4}{5}=\frac{2}{5}$. The denominator is $\frac{23}{30}$.

Dividing these, we get:

And there it is. We have successfully traced the effect back to its cause. The final probability is $\frac{12}{23}$.