Understanding the Probability Problem

This problem involves conditional probability. We have two bags, Bag A and Bag B, each containing a specific mix of red and black balls. We need to determine the likelihood that a randomly drawn red ball originated from Bag B, given the information about the bags' contents and the outcome of the draw.

Bag Contents:

• Bag A: Contains 3 Red balls and 4 Black balls. Total balls = $3 + 4 = 7$.
• Bag B: Contains 5 Red balls and 6 Black balls. Total balls = $5 + 6 = 11$.

Defining Events and Probabilities

Let's define the events involved:

• $B_A$: The event that Bag A is chosen.
• $B_B$: The event that Bag B is chosen.
• $R$: The event that a Red ball is drawn.

Since one bag is chosen at random, the probability of choosing either bag is equal:

• $P(B_A) = \frac{1}{2}$
• $P(B_B) = \frac{1}{2}$

Next, let's find the probability of drawing a red ball from each specific bag:

• The probability of drawing a red ball given Bag A was chosen is $P(R|B_A) = \frac{\text{Number of Red balls in Bag A}}{\text{Total balls in Bag A}} = \frac{3}{7}$.
• The probability of drawing a red ball given Bag B was chosen is $P(R|B_B) = \frac{\text{Number of Red balls in Bag B}}{\text{Total balls in Bag B}} = \frac{5}{11}$.

Applying Bayes' Theorem

We are asked to find the probability that the ball was drawn from Bag B, given that it is red. This is represented as $P(B_B|R)$. We can use Bayes' Theorem for this:

$P(B_B|R) = \frac{P(R|B_B) \times P(B_B)}{P(R)}$

First, we need to calculate the total probability of drawing a red ball, $P(R)$. This is done using the law of total probability:

$P(R) = P(R|B_A) \times P(B_A) + P(R|B_B) \times P(B_B)$

Substituting the values we found:

$P(R) = \left(\frac{3}{7} \times \frac{1}{2}\right) + \left(\frac{5}{11} \times \frac{1}{2}\right)$

$P(R) = \frac{3}{14} + \frac{5}{22}$

To add these fractions, we find a common denominator, which is 154:

$P(R) = \frac{3 \times 11}{14 \times 11} + \frac{5 \times 7}{22 \times 7}$

$P(R) = \frac{33}{154} + \frac{35}{154}$

$P(R) = \frac{33 + 35}{154} = \frac{68}{154}$

We can simplify this fraction by dividing both numerator and denominator by 2:

$P(R) = \frac{34}{77}$

Calculating the Final Probability

Now we can substitute the value of $P(R)$ back into Bayes' Theorem:

$P(B_B|R) = \frac{P(R|B_B) \times P(B_B)}{P(R)}$

$P(B_B|R) = \frac{\frac{5}{11} \times \frac{1}{2}}{\frac{34}{77}}$

$P(B_B|R) = \frac{\frac{5}{22}}{\frac{34}{77}}$

To divide the fractions, we multiply by the reciprocal of the denominator:

$P(B_B|R) = \frac{5}{22} \times \frac{77}{34}$

$P(B_B|R) = \frac{5 \times 77}{22 \times 34}$

We can simplify this expression by canceling common factors. Note that $77 = 7 \times 11$ and $22 = 2 \times 11$.

$P(B_B|R) = \frac{5 \times (7 \times 11)}{(2 \times 11) \times 34}$

Cancel out the 11:

$P(B_B|R) = \frac{5 \times 7}{2 \times 34}$

$P(B_B|R) = \frac{35}{68}$

Conclusion

The probability that the red ball was drawn from Bag B is $\frac{35}{68}$.