A bag contains N balls out of which 3 balls are white, 6 balls are green, and the remaining balls ar

Q: Wait, why do we subtract 2 from the denominator in the conditional probability?

Great observation! Because we are drawing 'without replacement', every time we take a ball out, the total count in the bag decreases. We drew one white ball and one green ball in the first two steps, so the total number of balls, which started at $N$, is now $N - 2$. We must account for this shift in the sample space.

Q: Why is the number of blue balls still $N-9$ in the third draw?

Excellent question. The condition given is $P(B_3 | W_1 \cap G_2)$. This means we are assuming that the first ball was white and the second was green. Since no blue balls were removed in those first two draws, the count of blue balls remains exactly what it was at the start: $N - (3 + 6) = N - 9$.

Q: Can I solve this using the $P(W_1 \cap G_2 \cap B_3) = \frac{2}{5N}$ equation instead?

You absolutely can! That is a fantastic way to verify your work. However, using the conditional probability $P(B_3 | W_1 \cap G_2) = \frac{2}{9}$ is a much faster, more direct path to finding $N$. Both methods will lead you to the same result, $N = 11$, which is the beauty of consistent mathematical logic.

Question

A bag contains $N$ balls out of which 3 balls are white, 6 balls are green, and the remaining balls are blue. Assume that the balls are identical otherwise. Three balls are drawn randomly one after the other without replacement. For $i = 1, 2, 3$, let $W_i, G_i$, and $B_i$ denote the events that the ball drawn in the $i^{th}$ draw is a white ball, green ball, and blue ball, respectively. If the probability $P(W_1 \cap G_2 \cap B_3) = \frac{2}{5N}$ and the conditional probability $P(B_3 | W_1 \cap G_2) = \frac{2}{9}$, then $N$ equals ________.

Sigma Percentile · Accepted Answer

### The Mystery of the Bag: A Journey into Conditional Probability

Imagine you are standing before a mysterious bag. Inside, there is a collection of balls—white, green, and blue—all identical to the touch, hiding their colors from you.

You are tasked with determining the total number of balls, $N$, based on the subtle clues of probability. This isn't just a math problem; it is a detective story where every draw changes the landscape of the bag.

### Phase 1: The Inventory

First, let us establish our baseline. We know the bag contains $N$ balls in total. We are given the counts for two of the colors: $3$ white balls and $6$ green balls.

The remaining balls are blue. To find the number of blue balls, we simply subtract the known quantities from the total:

$$ 	ext{Blue balls} = N - (3 + 6) = N - 9 $$

This expression, $N - 9$, is our anchor. It represents the population of blue balls that remains untouched as long as we do not draw one.

### Phase 2: The Conditional Shift

Now, we enter the realm of conditional probability. We are told that three balls are drawn without replacement. We focus on the event $P(B_3 | W_1 \cap G_2) = \frac{2}{9}$.

This notation is the language of the detective. It asks: 'If we know for a fact that the first ball was white and the second was green, what is the probability that the third ball is blue?'

Because we are drawing without replacement, the bag is a dynamic system. When we draw a white ball, the total count drops to $N - 1$. When we draw a green ball next, the total count drops again to $N - 2$.

Crucially, because we have not yet drawn a blue ball, the number of blue balls available remains $N - 9$.

### Phase 3: The Algebraic Bridge

We can now define the probability of the third draw being blue as the ratio of favorable outcomes (the blue balls) to the total possible outcomes (the remaining balls in the bag):

$$ P(B_3 | W_1 \cap G_2) = \frac{N - 9}{N - 2} $$

We are given that this probability is $\frac{2}{9}$. Now, we set up our equation:

$$ \frac{N - 9}{N - 2} = \frac{2}{9} $$

This is where the magic happens. By cross-multiplying, we clear the fractions and bring the equation into a linear form:

$$ 9(N - 9) = 2(N - 2) $$

Expanding both sides, we get:

$$ 9N - 81 = 2N - 4 $$

### Phase 4: The Resolution

Now, we isolate $N$. Subtracting $2N$ from both sides and adding $81$ to both sides, we find:

$$ 7N = 77 $$

Dividing by $7$, we arrive at the elegant solution: **$N = 11$**.

### Phase 5: The Victory Lap

In mathematics, the final answer is only half the joy; the other half is verification. Let us check our result against the other piece of information provided: $P(W_1 \cap G_2 \cap B_3) = \frac{2}{5N}$.

Using our value $N = 11$, the probability of drawing white, then green, then blue is:

$$ P(W_1) \cdot P(G_2 | W_1) \cdot P(B_3 | W_1 \cap G_2) = \frac{3}{11} \cdot \frac{6}{10} \cdot \frac{2}{9} = \frac{36}{990} = \frac{2}{55} $$

And the given formula $\frac{2}{5N}$ becomes:

$$ \frac{2}{5(11)} = \frac{2}{55} $$

The two values match perfectly! We have successfully navigated the constraints of the bag and uncovered the truth. Remember, in probability, always respect the changing state of your system—it is the key to unlocking the most complex problems.

Enter Numerical Value:

Visualized Solution

Initial State of the Bag

Counting Blue Balls

The Conditional Probability

Bag After Two Draws

Formulating the Equation

Solving for $N$

Expanding the Equation

Final Calculation

Verification and Conclusion

The Sigma Insight: Conditional Probability

The Mystery of the Bag

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

Similar Questions

Two integers are selected at random from the set ${1, 2, \dots, 11}$ . Given that the sum of selected numbers is even, the conditional probability that both the numbers are even is :

It is given that the events $A$ and $B$ are such that $P (A) = 1/4, P (A ∣ B) = 1/2$ and $P (B ∣ A) = 2/3$ . Then $P (B)$ is

If $A$ and $B$ are two events such that $P (A) = \frac{1}{3}$ , $P (B) = \frac{1}{5}$ and $P (A \cup B) = \frac{1}{2}$ , then $P (A ∣ B^{'}) + P (B ∣ A^{'})$ is equal to

Let A, B and C be three events, which are pair-wise independent and $\overset{ˉ}{E}$ denotes the complement of an event E. If $P (A \cap B \cap C) = 0$ and $P (C) > 0$ , then $P [(\overset{ˉ}{A} \cap \overset{ˉ}{B}) ∣ C]$ is equal to :

If $A$ and $B$ are two events such that $P (A) = 0.7$ , $P (B) = 0.4$ and $P (A \cap \overset{ˉ}{B}) = 0.5$ , where $\overset{ˉ}{B}$ denotes the complement of $B$ , then $P (B ∣ (A \cup \overset{ˉ}{B}))$ is equal:-

Let $A$ and $B$ be two events such that $P (B ∣ A) = \frac{2}{5}$ , $P (A ∣ B) = \frac{1}{7}$ and $P (A \cap B) = \frac{1}{9}$ . Consider (S1) $P (A^{'} \cup B) = \frac{5}{6}$ , (S2) $P (A^{'} \cap B^{'}) = \frac{1}{18}$ . Then

Three numbers are chosen at random without replacement from ${1, 2, 3, ..., 8}$ . The probability that their minimum is 3, given that their maximum is 6, is

Assume that each born child is equally likely to be a boy or a girl. If two families have two children each, then the conditional probability that all children are girls given that at least two are girls is :

$A$ is targeting to $B, B$ and $C$ are targeting to $A$ . Probability of hitting the target by $A, B$ and $C$ are $2/3, 1/2$ and $1/3$ respectively. If $A$ is hit then find the probability that $B$ hits the target and $C$ does not.