A bag contains 6 balls. Two balls are drawn from it at random and both are found to be black. The pr

Q: Wait, why do we assume the prior probabilities $P(B_i)$ are all equal?

Great question! In the absence of any specific information about the contents of the bag, we invoke the 'Principle of Indifference'. This principle states that if we have no reason to prefer one hypothesis over another, we must assign them equal probabilities. Since we don't know if the bag is biased toward more black or more white balls, assuming each scenario is equally likely is the only logical starting point.

Q: Why do we start our hypotheses at $i=2$ and not $i=0$?

Think about the physical evidence! We drew two black balls. If the bag had zero or one black ball, it would be physically impossible to draw two black balls. Therefore, those scenarios have a likelihood of zero. Since we are conditioning on the event that we *did* draw two black balls, we only care about the scenarios where that event is possible.

Q: Can we just use conditional probability directly without Bayes' Theorem?

Bayes' Theorem *is* the formal framework for conditional probability in this context. While you might try to reason through it intuitively, Bayes' Theorem provides the rigorous structure to update our 'prior' belief (that all bag compositions are equally likely) with the 'evidence' (the two black balls) to reach the 'posterior' probability. It is the most robust way to ensure you don't miss any terms in the denominator.

Question

A bag contains 6 balls. Two balls are drawn from it at random and both are found to be black. The probability that the bag contains at least 5 black balls is

Options:

(A) $\frac{5}{7}$

(B) $\frac{2}{7}$

(C) $\frac{3}{7}$

(D) $\frac{5}{6}$

Sigma Percentile · Accepted Answer

### The Mystery of the Black Balls

Imagine you are standing in a room with a mysterious, opaque bag. You know it contains exactly six balls, but you have absolutely no idea how many are black and how many are white.

You reach in, pull out two balls, and to your surprise, both are black. This is the moment of truth where probability stops being just a set of abstract rules and starts being a detective story. We are going to use Bayes' Theorem to reverse-engineer the contents of that bag.

### Defining the Hypotheses

Since we have already pulled out two black balls, we know for a fact that the bag must have contained at least two black balls. It could not have had zero or one.

Our possible hypotheses, which we will call $B_i$, represent the event that the bag contains exactly $i$ black balls, where $i \in \{2, 3, 4, 5, 6\}$. These are our five mutually exclusive scenarios.

We are essentially asking: "Given that I saw this result, which of these five worlds is the most likely to be the one I am standing in?"

### The Prior Probability

Before we drew the balls, we had no information. This is where the Principle of Indifference comes in.

We assume each of these five scenarios is equally likely. Thus, the prior probability $P(B_i)$ for each $i$ is $\frac{1}{5}$. It is a clean, unbiased starting point for our investigation.

### The Likelihood

Now, we consider the evidence $E$: the act of drawing two black balls. If the bag has $i$ black balls, the probability of drawing two black balls is given by the hypergeometric distribution:

$$P(E|B_i) = \frac{\binom{i}{2}}{\binom{6}{2}}$$

Since $\binom{6}{2} = 15$, the likelihoods are $\frac{\binom{i}{2}}{15}$. This formula is the heart of our calculation and quantifies how "surprising" our evidence is under each hypothesis.

If $i=6$, the probability is $1$ (certainty). If $i=2$, the probability is only $\frac{1}{15}$.

### Bayes' Theorem

We want to find the probability that the bag contains at least five black balls, given the evidence. This is $P(B_5 \cup B_6 | E)$.

By Bayes' Theorem, this is:

$$P(B_5 \cup B_6 | E) = \frac{P(E|B_5)P(B_5) + P(E|B_6)P(B_6)}{\sum_{i=2}^6 P(E|B_i)P(B_i)}$$

Since all $P(B_i)$ are equal to $\frac{1}{5}$, they cancel out beautifully. This leaves us with the ratio of the sum of the favorable likelihoods to the sum of all likelihoods.

### The Calculation

Let's calculate the individual likelihoods:

$$P(E|B_2) = \frac{1}{15}, \quad P(E|B_3) = \frac{3}{15}, \quad P(E|B_4) = \frac{6}{15}, \quad P(E|B_5) = \frac{10}{15}, \quad P(E|B_6) = \frac{15}{15}$$

The sum of the numerator (the favorable cases) is:

$$\frac{10}{15} + \frac{15}{15} = \frac{25}{15}$$

The sum of the denominator (all possible cases) is:

$$\frac{1+3+6+10+15}{15} = \frac{35}{15}$$

The final ratio is:

$$\frac{25}{35} = \frac{5}{7}$$

### Conclusion

And there you have it! The probability is **$\frac{5}{7}$**.

It is a beautiful result that shows how we can update our beliefs with new evidence. We started with five equally likely possibilities, but the evidence of two black balls shifted the weight heavily toward the scenarios where the bag is packed with black balls.

This is the power of Bayesian thinking—it turns uncertainty into a precise, calculated insight.

Animated Solution for Mathematics - Probability: A bag contains 6 balls. Two balls are drawn from it at random and both are found to be black. The probability that the bag contains at least 5 black balls is

Select Answer:

Visualized Solution

The Setup

Defining Hypotheses $B_{i}$

Prior Probabilities $P (B_{i})$

The Evidence Event $E$

Likelihood Formula $P (E ∣ B_{i})$

Calculating Total Outcomes

Likelihoods for $i = 2, 3, 4$

Likelihoods for $i = 5, 6$

Applying Bayes' Theorem

Substituting the Values

Summing the Terms

Final Simplification

The Sigma Insight: Bayes' Theorem

The Mystery of the Black Balls

Defining the Hypotheses

The Prior Probability

The Likelihood

Bayes' Theorem

The Calculation

Conclusion

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