Question

Bag A contains 3 white and 4 red balls, bag B contains 4 white and 5 red balls, and bag C contains 5 white and 6 red balls. If one ball is drawn at random from each of these three bags, then the probability of getting one white and two red balls is:

• A. \( \frac{268}{693} \)
• B. \( \frac{310}{693} \)
• C. \( \frac{38}{99} \)
• D. \( \frac{26}{63} \)

Hint:

When calculating probability for multiple events, first calculate the total possible outcomes and then find the favorable outcomes by considering all cases.

Answer 1

The Correct Option is D

The total number of ways to choose one ball from each bag is: \[ 7 \times 9 \times 11 = 693. \] Next, we calculate the favorable outcomes of getting one white and two red balls. There are 3 cases: - One white from bag A, one red from bag B, and one red from bag C: \[ 3 \times 5 \times 6 = 90. \] - One white from bag B, one red from bag A, and one red from bag C: \[ 4 \times 4 \times 6 = 96. \] - One white from bag C, one red from bag A, and one red from bag B: \[ 5 \times 4 \times 5 = 100. \] Therefore, the total number of favorable outcomes is: \[ 90 + 96 + 100 = 286. \] Thus, the probability is: \[ \frac{286}{693} = \frac{26}{63}. \] Therefore, the correct answer is \( \frac{26}{63} \).

Answer 2

Given:
- Bag A: 3 white and 4 red balls (total 7)
- Bag B: 4 white and 5 red balls (total 9)
- Bag C: 5 white and 6 red balls (total 11)
One ball is drawn from each bag.
Find the probability of getting exactly one white ball and two red balls.

Step 1: Possible cases for one white and two red balls:
- White from A, Red from B, Red from C
- Red from A, White from B, Red from C
- Red from A, Red from B, White from C

Step 2: Calculate probabilities for each case:
1) White from A, Red from B, Red from C:
\[ \frac{3}{7} \times \frac{5}{9} \times \frac{6}{11} = \frac{90}{693} \] 2) Red from A, White from B, Red from C:
\[ \frac{4}{7} \times \frac{4}{9} \times \frac{6}{11} = \frac{96}{693} \] 3) Red from A, Red from B, White from C:
\[ \frac{4}{7} \times \frac{5}{9} \times \frac{5}{11} = \frac{100}{693} \]

Step 3: Sum all probabilities:
\[ \frac{90}{693} + \frac{96}{693} + \frac{100}{693} = \frac{286}{693} \] Simplify numerator and denominator by dividing both by 11:
\[ \frac{286/11}{693/11} = \frac{26}{63} \]

Therefore,
\[ \boxed{\frac{26}{63}} \]