Question

Bag P contains 6 red and 4 blue balls, and bag Q contains 5 red and 6 blue balls. A ball is transferred from bag P to bag Q and then a ball is drawn from bag Q. What is the probability that the ball drawn is blue?

• A. \( \frac{7}{15} \)
• B. \( \frac{8}{15} \)
• C. \( \frac{4}{19} \)
• D. \( \frac{8}{19} \)

Hint:

When dealing with probability involving multiple events, break it down into cases, calculate the probability for each case, and then add them together for the total probability.

Answer 1

The Correct Option is B

The total probability consists of two cases: 
1. A blue ball is transferred from bag P to bag Q. 2. A red ball is transferred from bag P to bag Q. 
Case 1: A blue ball is transferred from bag P to bag Q. 
- The probability of selecting a blue ball from bag P is: \[ P({blue from P}) = \frac{4}{10} = \frac{2}{5} \] 
- After transferring the blue ball, bag Q contains 5 red and 7 blue balls. 
The probability of drawing a blue ball from bag Q is: \[ P({blue from Q after blue transfer}) = \frac{7}{12} \] 
Thus, the total probability for case 1 is: 
\[ P({blue transfer and blue drawn}) = \frac{2}{5} \times \frac{7}{12} = \frac{14}{60} = \frac{7}{30} \] 
Case 2: A red ball is transferred from bag P to bag Q. 
- The probability of selecting a red ball from bag P is: \[ P({red from P}) = \frac{6}{10} = \frac{3}{5} \] 
- After transferring the red ball, bag Q contains 6 red and 6 blue balls. The probability of drawing a blue ball from bag Q is: 
\[ P({blue from Q after red transfer}) = \frac{6}{12} = \frac{1}{2} \] 
Thus, the total probability for 
case 2 is: \[ P({red transfer and blue drawn}) = \frac{3}{5} \times \frac{6}{12} = \frac{18}{60} = \frac{3}{10} \] 
Total probability. The total probability of drawing a blue ball is the sum of the probabilities from both cases: 
\[ P({blue drawn}) = \frac{7}{30} + \frac{18}{60} = \frac{7}{30} + \frac{3}{10} = \frac{7}{30} + \frac{9}{30} = \frac{16}{30} = \frac{8}{15} \] 
Thus, the correct answer is \( \frac{8}{15} \).