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13 May 2024, 10:12
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r: 0.20
b: 0.30
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
87% (01:42) correct
13%
(01:56)
wrong
based on 325
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The probability of picking a red ball, a blue ball and a white ball from a bag containing 10 balls is r, b and w, where r<b<w. The bag contains balls of only three colors and there are at least two balls of each color.
Select for r the probability of picking red ball and select for b the probability of picking blue ball. Make only two selections, one in each column.
Select for r the probability of picking red ball and select for b the probability of picking blue ball. Make only two selections, one in each column.
| r | b | |
| 0.10 | ||
| 0.20 | ||
| 0.25 | ||
| 0.30 | ||
| 0.33 | ||
| 0.40 |
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Official Answer
r: 0.20
b: 0.30
13 May 2024, 18:46
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The probability of picking a red ball, a blue ball and a white ball from a bag containing 10 balls is r, b and w, where r<b<w. The bag contains balls of only three colors and there are at least two balls of each color.
Select for r the probability of picking red ball and select for b the probability of picking blue ball. Make only two selections, one in each column.
The fact that the question is asking for the probability of picking a red ball and the probability of picking a blue ball clues us into the fact that there must be some way to tell how many balls of each color there are in the bag.
We see that there are 10 balls in the bag, and r<b<w.
Thus, if the numbers of balls of each color are R, B, and W, R<B<W, and R + B + W = 10.
Since the passage says that there are are at least two balls of each color, let's try R = 2. In that case, B is at least 3, and W is at least 4.
However, since 2 + 3 + 4 = 9, W must be 5.
After all, we can't make R or B larger without making W larger as well. So, R cannot be 3, since then B and W would be at least 4 and 5, for a total of 12. Also, B cannot be 4 since, in that case, even if R = 2, W is at least 5, for a total of 2 + 4 + 5 = 11.
So, the only values that work for R, B, and W, are 2, 3, and 5, which total to 10.
Thus, the probability of picking a red ball is 2/10 or 0.20, and the probability of picking a blue ball is 3/10 or 0.30.
0.10
0.20
0.25
0.30
0.33
0.40
Correct answer: 0.20, 0.30
Select for r the probability of picking red ball and select for b the probability of picking blue ball. Make only two selections, one in each column.
The fact that the question is asking for the probability of picking a red ball and the probability of picking a blue ball clues us into the fact that there must be some way to tell how many balls of each color there are in the bag.
We see that there are 10 balls in the bag, and r<b<w.
Thus, if the numbers of balls of each color are R, B, and W, R<B<W, and R + B + W = 10.
Since the passage says that there are are at least two balls of each color, let's try R = 2. In that case, B is at least 3, and W is at least 4.
However, since 2 + 3 + 4 = 9, W must be 5.
After all, we can't make R or B larger without making W larger as well. So, R cannot be 3, since then B and W would be at least 4 and 5, for a total of 12. Also, B cannot be 4 since, in that case, even if R = 2, W is at least 5, for a total of 2 + 4 + 5 = 11.
So, the only values that work for R, B, and W, are 2, 3, and 5, which total to 10.
Thus, the probability of picking a red ball is 2/10 or 0.20, and the probability of picking a blue ball is 3/10 or 0.30.
0.10
0.20
0.25
0.30
0.33
0.40
Correct answer: 0.20, 0.30
13 Dec 2024, 12:11
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1. This type of problem is where there isn't a lot of restrictions, however, there is only one answer. To figure this out, you have to look at the minimum or maximum situation - let's do minimum.
2. Let the number of red, blue, and white balls be R, B, and W, respectively. There are at least 2 balls of each color, so, the minimum case is when: \(2 \leq R < B < W\) is \(2 < 3 < 4\).
3. In this case, there are 9 balls already. The 10th ball must be red, otherwise there is no way to split 10 balls into 3 colors so that they match the criterium.
4. That means R = 2, B = 3, and W = 5. Then the probabilities are: r = 2/10 = 0.2, b = 3/10 = 0.3, and w = 5/10 = 0.5.
5. Our answer will be: r - 0.2 and b - 0.3.
2. Let the number of red, blue, and white balls be R, B, and W, respectively. There are at least 2 balls of each color, so, the minimum case is when: \(2 \leq R < B < W\) is \(2 < 3 < 4\).
3. In this case, there are 9 balls already. The 10th ball must be red, otherwise there is no way to split 10 balls into 3 colors so that they match the criterium.
4. That means R = 2, B = 3, and W = 5. Then the probabilities are: r = 2/10 = 0.2, b = 3/10 = 0.3, and w = 5/10 = 0.5.
5. Our answer will be: r - 0.2 and b - 0.3.
