Aag A contains 4 Blue and 5 Yellow counters. Bag B contains 3 Blue and 6 Yellow
ounters. A counter is picked at random from Bag A and placed in Bag B. A
ounter is then picked at random from Bag B. Find the probability that the

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Let $$ A $$ contain 4 blue and 5 yellow counters, so there are a total of 9 counters. Bag $$ B $$ initially contains 3 blue and 6 yellow counters, a total of 9 counters.

We perform the experiment in two steps:

1. **Transfer a counter from Bag $$ A $$ to Bag $$ B $$.**

- The probability of transferring a blue counter from $$ A $$ is
     $$
     P(\text{Blue from } A)=\frac{4}{9}.
     $$
   - The probability of transferring a yellow counter is
     $$
     P(\text{Yellow from } A)=\frac{5}{9}.
     $$

2. **Draw a counter from Bag $$ B $$ after the transfer.**

When a counter is transferred from $$ A $$ to $$ B $$, the composition of Bag $$ B $$ changes:

- **If a blue counter is transferred:**
     
     Bag $$ B $$ then contains:
     - Blue: $$3+1=4$$
     - Yellow: $$6$$
     
     Total counters: $$10$$. Hence, the probability of drawing a blue counter from $$ B $$ is
     $$
     P(\text{Blue from } B \mid \text{Blue transferred})=\frac{4}{10}=\frac{2}{5}.
     $$

- **If a yellow counter is transferred:**
     
     Bag $$ B $$ then contains:
     - Blue: $$3$$
     - Yellow: $$6+1=7$$
     
     Total counters: $$10$$. Hence, the probability of drawing a blue counter from $$ B $$ is
     $$
     P(\text{Blue from } B \mid \text{Yellow transferred})=\frac{3}{10}.
     $$

3. **Calculate the total probability of drawing a blue counter from Bag $$ B $$:**

Using the law of total probability:
   $$
   \begin{aligned}
   P(\text{Blue drawn from } B) &= P(\text{Blue transferred}) \times P(\text{Blue from } B \mid \text{Blue transferred}) \
   &\quad + P(\text{Yellow transferred}) \times P(\text{Blue from } B \mid \text{Yellow transferred}) \
   &= \frac{4}{9}\times\frac{2}{5} + \frac{5}{9}\times\frac{3}{10}.
   \end{aligned}
   $$

Calculate each term:

- First term:
     $$
     \frac{4}{9}\times\frac{2}{5} = \frac{8}{45}.
     $$
   - Second term:
     $$
     \frac{5}{9}\times\frac{3}{10} = \frac{15}{90} = \frac{1}{6}.
     $$

Now, add the two results:
   $$
   P(\text{Blue drawn from } B) = \frac{8}{45}+\frac{1}{6}.
   $$

Find a common denominator, noting that the least common denominator of 45 and 6 is 90:
   $$
   \frac{8}{45} = \frac{16}{90} \quad \text{and} \quad \frac{1}{6} = \frac{15}{90}.
   $$
   Thus,
   $$
   P(\text{Blue drawn from } B) = \frac{16}{90}+\frac{15}{90}=\frac{31}{90}.
   $$

The probability that the counter picked from Bag $$ B $$ is blue is $$\boxed{\frac{31}{90}}$$.
The probability of drawing a blue counter from Bag B is $$\frac{31}{90}$$.

Aag A contains 4 Blue and 5 Yellow counters. Bag B contains 3 Blue and 6 Yellow
ounters. A counter is picked at random from Bag A and placed in Bag B. A
ounter is then picked at random from Bag B. Find the probability that the

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Aag A contains 4 Blue and 5 Yellow counters. Bag B contains 3 Blue and 6 Yellow ounters. A counter is picked at random from Bag A and placed in Bag B. A ounter is then picked at random from Bag B. Find the probability that the

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