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24 Jul 2023, 08:00
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Two boxes, A and B, each contains only red and blue balls. If the ratio of red to blue balls in box A is 3 to 2, what is the probability of drawing red ball from box B ?
(1) The probability of drawing one blue ball from box A and one blue ball from box B is 3/10
(2) The probability of drawing at least one red ball, when drawing one ball from box A and one ball from box B, is 7/10
(1) The probability of drawing one blue ball from box A and one blue ball from box B is 3/10
(2) The probability of drawing at least one red ball, when drawing one ball from box A and one ball from box B, is 7/10
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25 Apr 2025, 02:05
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Official Solution:
Two boxes, A and B, each contains only red and blue balls. If the ratio of red to blue balls in box A is 3 to 2, what is the probability of drawing red ball from box B ?
(1) The probability of drawing one blue ball from box A and one blue ball from box B is \(\frac{3}{10}\).
Since the ratio of red to blue balls in box A is 3 to 2, the probability of drawing a blue ball from box A is \(\frac{2}{5}\). Assuming the probability of drawing a blue ball from box B is \(p\), we have \(\frac{2}{5} * p = \frac{3}{10}\). Solving gives \(p = \frac{3}{4}\). Therefore, the probability of drawing a red ball from box B is \(1 - \frac{3}{4} = \frac{1}{4}\). Sufficient.
(2) The probability of drawing at least one red ball, when drawing one ball from box A and one ball from box B, is \(\frac{7}{10}\).
Assuming the probability of drawing a blue ball from box B is \(p\), we have \(1 - \text{(probability of drawing two blue balls)} = \frac{7}{10}\). Thus, \(1 - \frac{2}{5} * p = \frac{7}{10}\). Solving gives \(p = \frac{3}{4}\). Therefore, the probability of drawing a red ball from box B is \(1 - \frac{3}{4} = \frac{1}{4}\). Sufficient.
Answer: D
Two boxes, A and B, each contains only red and blue balls. If the ratio of red to blue balls in box A is 3 to 2, what is the probability of drawing red ball from box B ?
(1) The probability of drawing one blue ball from box A and one blue ball from box B is \(\frac{3}{10}\).
Since the ratio of red to blue balls in box A is 3 to 2, the probability of drawing a blue ball from box A is \(\frac{2}{5}\). Assuming the probability of drawing a blue ball from box B is \(p\), we have \(\frac{2}{5} * p = \frac{3}{10}\). Solving gives \(p = \frac{3}{4}\). Therefore, the probability of drawing a red ball from box B is \(1 - \frac{3}{4} = \frac{1}{4}\). Sufficient.
(2) The probability of drawing at least one red ball, when drawing one ball from box A and one ball from box B, is \(\frac{7}{10}\).
Assuming the probability of drawing a blue ball from box B is \(p\), we have \(1 - \text{(probability of drawing two blue balls)} = \frac{7}{10}\). Thus, \(1 - \frac{2}{5} * p = \frac{7}{10}\). Solving gives \(p = \frac{3}{4}\). Therefore, the probability of drawing a red ball from box B is \(1 - \frac{3}{4} = \frac{1}{4}\). Sufficient.
Answer: D
General Discussion
24 Jul 2023, 08:37
Kudos
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BOXES---A-----B
RED-----3---x
BLUE----2---y
probability of drawing red ball from box B : x/ ( x+y)
#1
The probability of drawing one blue ball from box A and one blue ball from box B is 3/10
2/5 * ( y/ (x+y)) = 3/10
4y =3x+3y
y=3x
BOXES---A-----B
RED-----3---3
BLUE----2---1
3/4 sufficient
#2
The probability of drawing at least one red ball, when drawing one ball from box A and one ball from box B, is 7/10
BOXES---A-----B
RED-----3---x
BLUE----2---y
three cases possible Red , red ; Red , blue ; Blue ,red
3/5 *(x/(x+y)) + 3/5 * ( (y/x+y) ) + 2/5 * (( x/(x+y) ) = 7/10
solve we get
3x=y
BOXES---A-----B
RED-----3---3
BLUE----2---1
3/4 P of Red ball box B
sufficient
OPTION D is correct
RED-----3---x
BLUE----2---y
probability of drawing red ball from box B : x/ ( x+y)
#1
The probability of drawing one blue ball from box A and one blue ball from box B is 3/10
2/5 * ( y/ (x+y)) = 3/10
4y =3x+3y
y=3x
BOXES---A-----B
RED-----3---3
BLUE----2---1
3/4 sufficient
#2
The probability of drawing at least one red ball, when drawing one ball from box A and one ball from box B, is 7/10
BOXES---A-----B
RED-----3---x
BLUE----2---y
three cases possible Red , red ; Red , blue ; Blue ,red
3/5 *(x/(x+y)) + 3/5 * ( (y/x+y) ) + 2/5 * (( x/(x+y) ) = 7/10
solve we get
3x=y
BOXES---A-----B
RED-----3---3
BLUE----2---1
3/4 P of Red ball box B
sufficient
OPTION D is correct
Bunuel
