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 \(P(W_1∩G_2∩B_3)=\frac{2}{5N}\) and the conditional probability \(P(B_3|W_1∩G_2)=\frac{2}{9}\), then N equals ________.

Answer 1

Correct Answer: 11

To solve the problem, use the given probabilities and conditional probability to find \(N\).

Given:
- Total balls = \(N\)
- White balls = 3
- Green balls = 6
- Blue balls = \(N - 9\) (since 3 + 6 = 9)
- Events:
\(W_i\) = \(i^{th}\) ball drawn is white
\(G_i\) = \(i^{th}\) ball drawn is green
\(B_i\) = \(i^{th}\) ball drawn is blue
- \(P(W_1 \cap G_2 \cap B_3) = \frac{2}{5N}\)
- \(P(B_3 | W_1 \cap G_2) = \frac{2}{9}\)

Step 1: Write expression for \(P(W_1 \cap G_2 \cap B_3)\)
Since drawing is without replacement: \[ P(W_1 \cap G_2 \cap B_3) = P(W_1) \times P(G_2 | W_1) \times P(B_3 | W_1 \cap G_2) \] Calculate each term: \[ P(W_1) = \frac{3}{N} \] After drawing one white ball, remaining balls = \(N-1\), green balls = 6: \[ P(G_2 | W_1) = \frac{6}{N - 1} \] Given: \[ P(B_3 | W_1 \cap G_2) = \frac{2}{9} \] So, \[ P(W_1 \cap G_2 \cap B_3) = \frac{3}{N} \times \frac{6}{N-1} \times \frac{2}{9} = \frac{36}{9 N (N-1)} = \frac{4}{N (N-1)} \]

Step 2: Equate this to given value \(\frac{2}{5 N}\):
\[ \frac{4}{N (N - 1)} = \frac{2}{5 N} \] Multiply both sides by \(N (N-1) 5 N\): \[ 4 \times 5 N = 2 \times (N)(N - 1) \] \[ 20 N = 2 N (N - 1) \] Divide both sides by \(2 N\) (assuming \(N \neq 0\)): \[ 10 = N - 1 \] \[ N = 11 \]

Final Answer:
\[ \boxed{11} \]

Answer 2

To solve the problem, we need to find the value of $N$, the total number of balls in the collection.

1. Given Data:
We are given that $W = 3$, $G = 6$, and $B = N - 9$ where $N$ is the total number of balls.

2. Probability Calculation for $W_1 \cap G_2 \cap B_3$:
The probability of drawing a white ball first, a green ball second, and a blue ball third is given by:

$ P(W_1 \cap G_2 \cap B_3) = P(W_1) P(G_2 | W_1) P(B_3 | W_1 \cap G_2) $

Substituting the values, we get:

$ P(W_1 \cap G_2 \cap B_3) = \frac{3}{N} \times \frac{6}{N-1} \times \frac{N-9}{N-2}$.

3. Setting Up the Equation:
We are given that $ P(W_1 \cap G_2 \cap B_3) = \frac{2}{5N} $, so we equate the two expressions:

$ \frac{3}{N} \times \frac{6}{N-1} \times \frac{N-9}{N-2} = \frac{2}{5N}$.

4. Solving for $N$:
Multiply both sides of the equation to get:

$ \frac{18(N-9)}{N(N-1)(N-2)} = \frac{2}{5N}$.
Now, multiply both sides by $5N$:

$ 9(N-9) \cdot 5N = N(N-1)(N-2)$.

5. Simplifying the Equation:
Expanding the terms, we get:

$ 45(N-9) = N^2 - 3N + 2$.
Now simplify further:

$ 45N - 405 = N^2 - 3N + 2$.
Rearranging the terms, we get:

$ N^2 - 48N + 407 = 0$.

6. Verifying $P(B_3 | W_1 \cap G_2)$:
Next, we are given that $ P(B_3 | W_1 \cap G_2) = \frac{2}{9}$. We calculate $ P(W_1 \cap G_2)$ as:

$ P(W_1 \cap G_2) = P(W_1) P(G_2 | W_1) = \frac{3}{N} \times \frac{6}{N-1} = \frac{18}{N(N-1)}$.
Substituting into the equation for $P(B_3 | W_1 \cap G_2)$, we get:

$ P(B_3 | W_1 \cap G_2) = \frac{\frac{3}{N} \times \frac{6}{N-1} \times \frac{N-9}{N-2}}{\frac{3}{N} \times \frac{6}{N-1}} = \frac{N-9}{N-2}$.
We are given that $ P(B_3 | W_1 \cap G_2) = \frac{2}{9}$, so:

$ 9(N-9) = 2(N-2)$.
Expanding both sides gives:

$ 9N - 81 = 2N - 4$.
Solving for $N$ gives:

$ 7N = 77$ and $N = 11$.

7. Final Verification:
Substituting $N = 11$ into the equation for $P(W_1 \cap G_2 \cap B_3)$:

$ P(W_1 \cap G_2 \cap B_3) = \frac{3}{11} \times \frac{6}{10} \times \frac{2}{9} = \frac{36}{990} = \frac{18}{495} = \frac{2}{55}$.
Since $P(W_1 \cap G_2 \cap B_3) = \frac{2}{5N}$, we verify that $ \frac{2}{55} = \frac{2}{5N}$, so $N = 11$ is correct.

Final Answer:
The final answer is $\boxed{11}$.