There are two bags:
- Bag 1: 4 red, 3 black → total = 7 balls
- Bag 2: 2 red, 3 black → total = 5 balls
A bag is chosen at random and one ball is drawn. What is the probability that the ball is red?
- Bag 1: 4 red, 3 black → total = 7 balls
- Bag 2: 2 red, 3 black → total = 5 balls
A bag is chosen at random and one ball is drawn. What is the probability that the ball is red?
Show Hint
- \( \frac{39}{70} \)
- \( \frac{41}{70} \)
- \( \frac{29}{70} \)
- \( \frac{17}{35} \)
The Correct Option is D
Solution and Explanation
Let:
- \( R \): drawing a red ball
- \( B_1 \): choosing Bag 1
- \( B_2 \): choosing Bag 2
\[ P(B_1) = P(B_2) = \frac{1}{2} \] - \( P(R \mid B_1) = \frac{4}{7} \)
- \( P(R \mid B_2) = \frac{2}{5} \)
Then total probability: \[ P(R) = P(B_1) \cdot P(R \mid B_1) + P(B_2) \cdot P(R \mid B_2) = \frac{1}{2} \cdot \frac{4}{7} + \frac{1}{2} \cdot \frac{2}{5} = \frac{2}{7} + \frac{1}{5} \] Take LCM: \[ = \frac{10 + 7}{35} = \frac{17}{35} \]
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