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Sagot :
To determine [tex]\(P(B \mid A)\)[/tex], the probability of drawing a white ball on the second draw given that the first draw was a blue ball, follow these steps:
1. Identify the total number of balls:
There are 4 blue balls, 7 yellow balls, and 4 white balls. The total number of balls in the bag is:
[tex]\[ 4 + 7 + 4 = 15 \][/tex]
2. Calculate the probability of drawing a blue ball first (Event A):
The probability of drawing one of the 4 blue balls from the 15 balls is:
[tex]\[ P(A) = \frac{4}{15} \][/tex]
3. Adjust the counts after drawing the first ball:
If a blue ball is drawn first, there are now 14 balls left in the bag (since the ball is not replaced). The counts are now:
- Blue balls: 3
- Yellow balls: 7
- White balls: 4
4. Determine the probability of drawing a white ball second given a blue ball was drawn first (Event B given A):
Now, with 14 balls left in the bag, the probability of drawing one of the 4 white balls is:
[tex]\[ P(B \mid A) = \frac{4}{14} \][/tex]
5. Simplify the fraction:
[tex]\[ P(B \mid A) = \frac{4}{14} = \frac{2}{7} \][/tex]
Therefore, the probability of drawing a white ball on the second draw given that the first draw was a blue ball, [tex]\(P(B \mid A)\)[/tex], is:
[tex]\[ \boxed{\frac{2}{7}} \][/tex]
1. Identify the total number of balls:
There are 4 blue balls, 7 yellow balls, and 4 white balls. The total number of balls in the bag is:
[tex]\[ 4 + 7 + 4 = 15 \][/tex]
2. Calculate the probability of drawing a blue ball first (Event A):
The probability of drawing one of the 4 blue balls from the 15 balls is:
[tex]\[ P(A) = \frac{4}{15} \][/tex]
3. Adjust the counts after drawing the first ball:
If a blue ball is drawn first, there are now 14 balls left in the bag (since the ball is not replaced). The counts are now:
- Blue balls: 3
- Yellow balls: 7
- White balls: 4
4. Determine the probability of drawing a white ball second given a blue ball was drawn first (Event B given A):
Now, with 14 balls left in the bag, the probability of drawing one of the 4 white balls is:
[tex]\[ P(B \mid A) = \frac{4}{14} \][/tex]
5. Simplify the fraction:
[tex]\[ P(B \mid A) = \frac{4}{14} = \frac{2}{7} \][/tex]
Therefore, the probability of drawing a white ball on the second draw given that the first draw was a blue ball, [tex]\(P(B \mid A)\)[/tex], is:
[tex]\[ \boxed{\frac{2}{7}} \][/tex]
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