Question

Bag-I contains 1 white, 2 Red and 3 Blue Balls. Bag-II contains 2 white, 3 Red and 4 Blue Balls. An experiment involves picking 3 balls from Bag-I without noticing its color and placing it in Bag-II. If it is known that there are atleast 5 Blue Balls in Bag-II after the experiment, then the probability that there exists atmost 1 red ball in the three picked balls is $\lambda$, then $190\lambda$ = ______

Answer 1

160

Answer 2

Let $B_1$ be Bag-I and $B_2$ be Bag-II.

Bag-I contains 1 White (W), 2 Red (R), and 3 Blue (B) balls. Total balls in $B_1$ = 1 + 2 + 3 = 6.

Bag-II contains 2 White (W), 3 Red (R), and 4 Blue (B) balls. Total balls in $B_2$ = 2 + 3 + 4 = 9.

We pick 3 balls from Bag-I and place them in Bag-II. The total number of ways to pick 3 balls from the 6 balls in Bag-I is $\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.

Let $(n_W, n_R, n_B)$ denote the number of White, Red, and Blue balls picked from Bag-I, such that $n_W + n_R + n_B = 3$. The number of ways to pick such a combination is $\binom{1}{n_W}\binom{2}{n_R}\binom{3}{n_B}$.

The possible compositions and the number of ways are:

• (0, 0, 3): $\binom{1}{0}\binom{2}{0}\binom{3}{3} = 1 \times 1 \times 1 = 1$ way.
• (0, 1, 2): $\binom{1}{0}\binom{2}{1}\binom{3}{2} = 1 \times 2 \times 3 = 6$ ways.
• (0, 2, 1): $\binom{1}{0}\binom{2}{2}\binom{3}{1} = 1 \times 1 \times 3 = 3$ ways.
• (1, 0, 2): $\binom{1}{1}\binom{2}{0}\binom{3}{2} = 1 \times 1 \times 3 = 3$ ways.
• (1, 1, 1): $\binom{1}{1}\binom{2}{1}\binom{3}{1} = 1 \times 2 \times 3 = 6$ ways.
• (1, 2, 0): $\binom{1}{1}\binom{2}{2}\binom{3}{0} = 1 \times 1 \times 1 = 1$ way.

The total number of ways is $1+6+3+3+6+1 = 20$, which matches $\binom{6}{3}$.

Let $E$ be the event that there are at least 5 Blue balls in Bag-II after the experiment.

Initially, there are 4 Blue balls in Bag-II. After adding $n_B$ Blue balls from Bag-I, the number of Blue balls in Bag-II is $4 + n_B$.

The condition $4 + n_B \ge 5$ implies $n_B \ge 1$.

The compositions satisfying $n_B \ge 1$ are those with $n_B=1, 2,$ or $3$: (0, 0, 3), (0, 1, 2), (0, 2, 1), (1, 0, 2), (1, 1, 1).

The total number of ways for event $E$ is $1 + 6 + 3 + 3 + 6 = 19$.

The probability of event $E$ is $P(E) = \frac{\text{Number of ways for E}}{\text{Total number of ways}} = \frac{19}{20}$.

Let $A$ be the event that there exists at most 1 red ball in the three picked balls from Bag-I.

This means $n_R \le 1$.

The compositions satisfying $n_R \le 1$ are those with $n_R=0$ or $n_R=1$: (0, 0, 3), (1, 0, 2), (0, 1, 2), (1, 1, 1).

The total number of ways for event $A$ is $1 + 3 + 6 + 6 = 16$.

The probability of event $A$ is $P(A) = \frac{16}{20} = \frac{4}{5}$.

We are interested in the probability of event $A$ given event $E$, which is $P(A|E) = \frac{P(A \cap E)}{P(E)}$.

The event $A \cap E$ is that the picked balls satisfy both $n_R \le 1$ and $n_B \ge 1$.

Let's check the compositions satisfying $n_R \le 1$:

• (0, 0, 3): $n_R=0, n_B=3$. $n_R \le 1$ (Yes), $n_B \ge 1$ (Yes). In $A \cap E$.
• (1, 0, 2): $n_R=0, n_B=2$. $n_R \le 1$ (Yes), $n_B \ge 1$ (Yes). In $A \cap E$.
• (0, 1, 2): $n_R=1, n_B=2$. $n_R \le 1$ (Yes), $n_B \ge 1$ (Yes). In $A \cap E$.
• (1, 1, 1): $n_R=1, n_B=1$. $n_R \le 1$ (Yes), $n_B \ge 1$ (Yes). In $A \cap E$.

The compositions satisfying $n_R \le 1$ but not $n_B \ge 1$ would be those with $n_R \le 1$ and $n_B = 0$. There are no such compositions summing to 3 with the given balls in Bag-I (e.g., (1, 1, 0) is not possible as $n_W+n_R+n_B=1+1+0=2 \ne 3$).

The compositions satisfying $n_B \ge 1$ but not $n_R \le 1$ would be those with $n_B \ge 1$ and $n_R > 1$. This means $n_R=2$.

• (0, 2, 1): $n_R=2, n_B=1$. $n_R \le 1$ (No). Not in $A \cap E$.

The compositions satisfying both $n_R \le 1$ and $n_B \ge 1$ are (0, 0, 3), (1, 0, 2), (0, 1, 2), (1, 1, 1).

The number of ways for $A \cap E$ is $1 + 3 + 6 + 6 = 16$.

The probability of $A \cap E$ is $P(A \cap E) = \frac{16}{20} = \frac{4}{5}$.

The conditional probability $P(A|E)$ is: $P(A|E) = \frac{P(A \cap E)}{P(E)} = \frac{16/20}{19/20} = \frac{16}{19}$.

The question states that this probability is $\lambda$. So, $\lambda = \frac{16}{19}$.

We need to find the value of $190\lambda$. $190\lambda = 190 \times \frac{16}{19} = 10 \times 16 = 160$.

The final answer is $\boxed{160}$.