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A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag. Its colour is seen and this ball, along with two additional balls of the same colour, is returned to the bag. If now, a ball is drawn at random from the bag, then the probability that this drawn ball is red is:
- a)3/10
- b)2/5
- c)1/5
- d)3/4
Correct answer is option 'B'. Can you explain this answer?
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A bag contains 4 red and 6 black balls. A ball is drawn at random from...
E1: Event that first ball drawn is red
E2: Event that first ball drawn is black
E: Event that second ball drawn is red
E2: Event that first ball drawn is black
E: Event that second ball drawn is red
Most Upvoted Answer
A bag contains 4 red and 6 black balls. A ball is drawn at random from...
Understanding the Problem
We have a bag containing 4 red balls and 6 black balls. When a ball is drawn, its color is noted, and two additional balls of the same color are added back to the bag.
We need to calculate the probability that a ball drawn afterwards is red.
Initial Setup
- Total red balls: 4
- Total black balls: 6
- Total balls: 4 + 6 = 10
Drawing Scenarios
1. If a red ball is drawn:
- We have 4 red balls, so the probability of drawing a red ball first is:
\[ P(R) = \frac{4}{10} = \frac{2}{5} \]
- After drawing and replacing with 2 more red balls:
- New count: 6 red balls, 6 black balls.
- Total: 12 balls.
2. If a black ball is drawn:
- The probability of drawing a black ball first is:
\[ P(B) = \frac{6}{10} = \frac{3}{5} \]
- After drawing and replacing with 2 more black balls:
- New count: 4 red balls, 8 black balls.
- Total: 12 balls.
Calculating Final Probabilities
- Probability of drawing a red ball after drawing red first:
\[ P(R | R) = \frac{6}{12} = \frac{1}{2} \]
- Probability of drawing a red ball after drawing black first:
\[ P(R | B) = \frac{4}{12} = \frac{1}{3} \]
Total Probability of Drawing a Red Ball
Using the law of total probability:
\[
P(R) = P(R | R) \cdot P(R) + P(R | B) \cdot P(B)
\]
Substituting the values:
\[
P(R) = \left(\frac{1}{2} \cdot \frac{2}{5}\right) + \left(\frac{1}{3} \cdot \frac{3}{5}\right)
\]
Calculating:
\[
P(R) = \frac{1}{5} + \frac{1}{5} = \frac{2}{5}
\]
Conclusion
The probability that the drawn ball is red is:
\[
\text{Option B: } \frac{2}{5}
\]
We have a bag containing 4 red balls and 6 black balls. When a ball is drawn, its color is noted, and two additional balls of the same color are added back to the bag.
We need to calculate the probability that a ball drawn afterwards is red.
Initial Setup
- Total red balls: 4
- Total black balls: 6
- Total balls: 4 + 6 = 10
Drawing Scenarios
1. If a red ball is drawn:
- We have 4 red balls, so the probability of drawing a red ball first is:
\[ P(R) = \frac{4}{10} = \frac{2}{5} \]
- After drawing and replacing with 2 more red balls:
- New count: 6 red balls, 6 black balls.
- Total: 12 balls.
2. If a black ball is drawn:
- The probability of drawing a black ball first is:
\[ P(B) = \frac{6}{10} = \frac{3}{5} \]
- After drawing and replacing with 2 more black balls:
- New count: 4 red balls, 8 black balls.
- Total: 12 balls.
Calculating Final Probabilities
- Probability of drawing a red ball after drawing red first:
\[ P(R | R) = \frac{6}{12} = \frac{1}{2} \]
- Probability of drawing a red ball after drawing black first:
\[ P(R | B) = \frac{4}{12} = \frac{1}{3} \]
Total Probability of Drawing a Red Ball
Using the law of total probability:
\[
P(R) = P(R | R) \cdot P(R) + P(R | B) \cdot P(B)
\]
Substituting the values:
\[
P(R) = \left(\frac{1}{2} \cdot \frac{2}{5}\right) + \left(\frac{1}{3} \cdot \frac{3}{5}\right)
\]
Calculating:
\[
P(R) = \frac{1}{5} + \frac{1}{5} = \frac{2}{5}
\]
Conclusion
The probability that the drawn ball is red is:
\[
\text{Option B: } \frac{2}{5}
\]
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A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag. Its colour is seen and this ball, along with two additional balls of the same colour, is returned to the bag. If now, a ball is drawn at random from the bag, then the probability that this drawn ball is red is:a)3/10b)2/5c)1/5d)3/4Correct answer is option 'B'. Can you explain this answer?
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