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Sagot :
Let's analyze the problem step-by-step.
### Part (A): Probability that the second ball was red, given that the first ball was replaced before the second draw.
With replacement means that after drawing the first ball, it is put back into the box.
1. Initially, the box contains a total of 12 balls (5 red and 7 white). After drawing and replacing the first ball, the box still contains the same 12 balls.
2. The probability of drawing a red ball on any draw (first or second) is the same because after each draw the box composition remains unchanged.
3. Therefore, the probability of drawing a red ball on the second draw is the ratio of the number of red balls to the total number of balls.
[tex]\[ \text{Probability (second ball is red with replacement)} \ = \frac{5}{12} \approx 0.4167 \][/tex]
### Part (B): Probability that the second ball was red, given that the first ball was not replaced before the second draw.
Without replacement means that after drawing the first ball, it is not put back into the box.
1. Initially, the box contains 12 balls (5 red and 7 white).
Case 1: The first ball drawn is red.
- If the first ball is red, then there are 4 red balls remaining out of a total of 11 balls.
- The probability of drawing a red ball first is [tex]\( \frac{5}{12} \)[/tex].
- Given the first ball was red, the probability of drawing a red ball on the second draw is [tex]\( \frac{4}{11} \)[/tex].
Case 2: The first ball drawn is white.
- If the first ball is white, then there are still 5 red balls remaining out of a total of 11 balls.
- The probability of drawing a white ball first is [tex]\( \frac{7}{12} \)[/tex].
- Given the first ball was white, the probability of drawing a red ball on the second draw is [tex]\( \frac{5}{11} \)[/tex].
Now, to find the total probability of drawing a red ball on the second draw without replacement, we sum the probabilities of both cases:
[tex]\[ \text{Probability (first red, then red)} = \left( \frac{5}{12} \right) \times \left( \frac{4}{11} \right) \][/tex]
[tex]\[ \text{Probability (first white, then red)} = \left( \frac{7}{12} \right) \times \left( \frac{5}{11} \right) \][/tex]
[tex]\[ \text{Total probability} = \left( \frac{5}{12} \times \frac{4}{11} \right) + \left( \frac{7}{12} \times \frac{5}{11} \right) \][/tex]
[tex]\[ = \left( \frac{20}{132} \right) + \left( \frac{35}{132} \right) = \frac{55}{132} \approx 0.4167 \][/tex]
Therefore:
[tex]\[ \text{Probability (second ball is red without replacement)} \ \approx 0.4167 \][/tex]
### Summary:
(A) Probability that the second ball drawn was red, given that the first ball drawn was replaced before the second draw:
[tex]\[ \approx 0.4167 \][/tex]
(B) Probability that the second ball drawn was red, given that the first ball drawn was not replaced before the second draw:
[tex]\[ \approx 0.4167 \][/tex]
### Part (A): Probability that the second ball was red, given that the first ball was replaced before the second draw.
With replacement means that after drawing the first ball, it is put back into the box.
1. Initially, the box contains a total of 12 balls (5 red and 7 white). After drawing and replacing the first ball, the box still contains the same 12 balls.
2. The probability of drawing a red ball on any draw (first or second) is the same because after each draw the box composition remains unchanged.
3. Therefore, the probability of drawing a red ball on the second draw is the ratio of the number of red balls to the total number of balls.
[tex]\[ \text{Probability (second ball is red with replacement)} \ = \frac{5}{12} \approx 0.4167 \][/tex]
### Part (B): Probability that the second ball was red, given that the first ball was not replaced before the second draw.
Without replacement means that after drawing the first ball, it is not put back into the box.
1. Initially, the box contains 12 balls (5 red and 7 white).
Case 1: The first ball drawn is red.
- If the first ball is red, then there are 4 red balls remaining out of a total of 11 balls.
- The probability of drawing a red ball first is [tex]\( \frac{5}{12} \)[/tex].
- Given the first ball was red, the probability of drawing a red ball on the second draw is [tex]\( \frac{4}{11} \)[/tex].
Case 2: The first ball drawn is white.
- If the first ball is white, then there are still 5 red balls remaining out of a total of 11 balls.
- The probability of drawing a white ball first is [tex]\( \frac{7}{12} \)[/tex].
- Given the first ball was white, the probability of drawing a red ball on the second draw is [tex]\( \frac{5}{11} \)[/tex].
Now, to find the total probability of drawing a red ball on the second draw without replacement, we sum the probabilities of both cases:
[tex]\[ \text{Probability (first red, then red)} = \left( \frac{5}{12} \right) \times \left( \frac{4}{11} \right) \][/tex]
[tex]\[ \text{Probability (first white, then red)} = \left( \frac{7}{12} \right) \times \left( \frac{5}{11} \right) \][/tex]
[tex]\[ \text{Total probability} = \left( \frac{5}{12} \times \frac{4}{11} \right) + \left( \frac{7}{12} \times \frac{5}{11} \right) \][/tex]
[tex]\[ = \left( \frac{20}{132} \right) + \left( \frac{35}{132} \right) = \frac{55}{132} \approx 0.4167 \][/tex]
Therefore:
[tex]\[ \text{Probability (second ball is red without replacement)} \ \approx 0.4167 \][/tex]
### Summary:
(A) Probability that the second ball drawn was red, given that the first ball drawn was replaced before the second draw:
[tex]\[ \approx 0.4167 \][/tex]
(B) Probability that the second ball drawn was red, given that the first ball drawn was not replaced before the second draw:
[tex]\[ \approx 0.4167 \][/tex]
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