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27 May 2025, 11:27
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A: \(\frac{2}{3}\)
B: \(\frac{1}{3}\)
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
32% (03:00) correct
68%
(03:05)
wrong
based on 157
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A monkey is playing a game with three empty cups, labeled 1, 2, and 3. The monkey has one red ball and one blue ball, and places them into the cups using the following rules:
- The red and blue balls are placed uniformly at random into one of the three cups.
- It is known that the monkey never puts the blue ball into cup 3.
- Each cup can hold more than one ball.
Let A be the probability that cup 1 contains at least one ball and let B be the probability that the red and blue balls end up in cups 1 and 2 (in either order).
- The red and blue balls are placed uniformly at random into one of the three cups.
- It is known that the monkey never puts the blue ball into cup 3.
- Each cup can hold more than one ball.
Let A be the probability that cup 1 contains at least one ball and let B be the probability that the red and blue balls end up in cups 1 and 2 (in either order).
| A | B | |
| \(\frac{1}{6}\) | ||
| \(\frac{1}{3}\) | ||
| \(\frac{1}{2}\) | ||
| \(\frac{2}{3}\) | ||
| \(\frac{3}{4}\) | ||
| \(\frac{5}{6}\) |
ShowHide Answer
Official Answer
A: \(\frac{2}{3}\)
B: \(\frac{1}{3}\)
27 May 2025, 18:14
Kudos
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A monkey is playing a game with three empty cups, labeled 1, 2, and 3. The monkey has one red ball and one blue ball, and places them into the cups using the following rules:
- The red and blue balls are placed uniformly at random into one of the three cups.
- It is known that the monkey never puts the blue ball into cup 3.
- Each cup can hold more than one ball.
Let A be the probability that cup 1 contains at least one ball and let B be the probability that the red and blue balls end up in cups 1 and 2 (in either order).
Calculate A
Probability that cup 1 contains at least one ball = 1 - Probability that cup 1 contains no balls
Probability that blue does not go into cup 1: 1/2
Probability that red does not go into cup 1: 2/3
1 - (1/2 × 2/3) = 1 - 1/3 = 2/3
For A, select 2/3.
Calculate B
We can start with the blue ball. The blue ball can go into either cup 1 or cup 2.
Probability that blue goes into cup 1 or cup 2: 1
Then, once blue is in a cup, there's a 1 in 3 chance that red will be in the other of cup 1 and cup 2.
For example, if blue is in cup 1, there's a 1 in 3 chance that red goes into cup 2.
1 × 1/3 = 1/3
For B, select 1/3.
Correct answer: 2/3, 1/3
- The red and blue balls are placed uniformly at random into one of the three cups.
- It is known that the monkey never puts the blue ball into cup 3.
- Each cup can hold more than one ball.
Let A be the probability that cup 1 contains at least one ball and let B be the probability that the red and blue balls end up in cups 1 and 2 (in either order).
Calculate A
Probability that cup 1 contains at least one ball = 1 - Probability that cup 1 contains no balls
Probability that blue does not go into cup 1: 1/2
Probability that red does not go into cup 1: 2/3
1 - (1/2 × 2/3) = 1 - 1/3 = 2/3
For A, select 2/3.
Calculate B
We can start with the blue ball. The blue ball can go into either cup 1 or cup 2.
Probability that blue goes into cup 1 or cup 2: 1
Then, once blue is in a cup, there's a 1 in 3 chance that red will be in the other of cup 1 and cup 2.
For example, if blue is in cup 1, there's a 1 in 3 chance that red goes into cup 2.
1 × 1/3 = 1/3
For B, select 1/3.
Correct answer: 2/3, 1/3
29 Jun 2025, 10:50
Kudos
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1. The monkey randomly places a red and blue ball in 3 cups with some restrictions. We are asked to find probabilities A and B of certain events.
2. Let's write down all the different possibilities:
3. Probability A is when the first cup has at least one ball, so \(A = \frac{4}{6} = \frac{2}{3}\). Probability B is when the red and blue balls take up the first two cups, so \(B = \frac{2}{6} = \frac{1}{3}\).
4. Our answer will be: A - \(\frac{2}{3}\) and B - \(\frac{1}{3}\).
2. Let's write down all the different possibilities:
| Cups: | 1 | 2 | 3 |
| R,B | |||
| R | B | ||
| B | R | ||
| R,B | |||
| B | R | ||
| B | R |
3. Probability A is when the first cup has at least one ball, so \(A = \frac{4}{6} = \frac{2}{3}\). Probability B is when the red and blue balls take up the first two cups, so \(B = \frac{2}{6} = \frac{1}{3}\).
4. Our answer will be: A - \(\frac{2}{3}\) and B - \(\frac{1}{3}\).
