Loading...
A box contains 11 balls, of which 6 are yellow and 5 are blue. If 2 balls are chosen at random without replacement, what is the probability that 1 will be yellow and 1 will be blue?
Let's break down what we're looking for in everyday terms. We have a box with 11 balls total - 6 yellow ones and 5 blue ones. We're going to pick 2 balls, one after the other, without putting the first ball back. We want to know: what are the chances that we end up with exactly one yellow ball and one blue ball?
Think of it like picking two different colored socks from a drawer in the dark. We want one of each color, not two of the same color.
Process Skill: TRANSLATE - Converting the problem language into a clear mathematical understanding of what we need to find
Now, let's think about the different ways we can get one yellow and one blue ball. Since we're picking them one at a time, there are exactly two ways this can happen:
That's it! These are the only two ways to end up with one ball of each color. Both scenarios give us exactly what we want - one yellow and one blue.
Let's work out the probability for each scenario step by step:
Scenario A: Yellow first, then Blue
• First pick (yellow): We have 6 yellow balls out of 11 total balls
Probability = \(\frac{6}{11}\)
• Second pick (blue): After removing one yellow ball, we now have 5 blue balls out of 10 remaining balls
Probability = \(\frac{5}{10} = \frac{1}{2}\)
• Probability of Scenario A = \(\left(\frac{6}{11}\right) \times \left(\frac{1}{2}\right) = \frac{6}{22} = \frac{3}{11}\)
Scenario B: Blue first, then Yellow
• First pick (blue): We have 5 blue balls out of 11 total balls
Probability = \(\frac{5}{11}\)
• Second pick (yellow): After removing one blue ball, we now have 6 yellow balls out of 10 remaining balls
Probability = \(\frac{6}{10} = \frac{3}{5}\)
• Probability of Scenario B = \(\left(\frac{5}{11}\right) \times \left(\frac{3}{5}\right) = \frac{15}{55} = \frac{3}{11}\)
Notice that both scenarios have the same probability: \(\frac{3}{11}\). This makes sense because we have a similar number of yellow and blue balls!
Since either scenario A or scenario B will give us what we want (one yellow and one blue), we add their probabilities together:
Total probability = Probability of Scenario A + Probability of Scenario B
Total probability = \(\frac{3}{11} + \frac{3}{11} = \frac{6}{11}\)
Technical notation: \(\mathrm{P}(1 \text{ yellow, } 1 \text{ blue}) = \mathrm{P}(\text{Y then B}) + \mathrm{P}(\text{B then Y}) = \frac{6}{11}\)
The probability of getting exactly one yellow ball and one blue ball is \(\frac{6}{11}\).
Looking at our answer choices, this matches choice D: \(\frac{6}{11}\).
We can verify this makes sense: \(\frac{6}{11}\) is slightly more than \(\frac{1}{2}\), which seems reasonable since we have slightly more yellow balls than blue balls, making it quite likely we'll get one of each color when picking two balls.
1. Misunderstanding "without replacement"
Students often overlook that "without replacement" means the first ball picked is NOT put back before picking the second ball. This changes the total number of balls and the composition for the second pick. Some students might incorrectly assume both picks are independent with the same probabilities.
2. Missing one of the two scenarios
Students frequently forget that there are TWO ways to get one yellow and one blue ball: (Yellow first, Blue second) OR (Blue first, Yellow second). They might only calculate one scenario and miss adding the probability of the other scenario, leading to an incomplete answer.
3. Confusing "exactly one of each" with other interpretations
Some students might misinterpret "1 will be yellow and 1 will be blue" as meaning "at least one yellow" or "at least one blue" rather than "exactly one of each color." This changes the entire approach to the problem.
1. Forgetting to update the denominator for the second pick
After picking the first ball, students often forget that there are now only 10 balls left (not 11) for the second pick. They might incorrectly use 11 as the denominator for both picks.
2. Forgetting to update the numerator for the second pick
Students might correctly reduce the total to 10 balls but forget to adjust the number of remaining balls of each color. For example, if a yellow ball is picked first, there are only 5 yellow balls left (not 6) for the second pick.
3. Arithmetic errors in fraction multiplication and addition
When calculating \(\left(\frac{6}{11}\right) \times \left(\frac{5}{10}\right)\) or \(\left(\frac{5}{11}\right) \times \left(\frac{6}{10}\right)\), students might make multiplication errors. When adding \(\frac{3}{11} + \frac{3}{11}\), some might forget basic addition rules or make computational mistakes.
1. Selecting only one scenario's probability instead of the sum
Students might correctly calculate both scenarios (\(\frac{3}{11}\) each) but accidentally select \(\frac{3}{11}\) (answer choice B) instead of adding them together to get \(\frac{6}{11}\). This happens when they lose track of needing to combine both scenarios.
2. Selecting a probability greater than 1 or an unreasonable answer
Some students might make calculation errors that lead to impossible probabilities. They should check that their answer makes intuitive sense - \(\frac{6}{11}\) (slightly more than half) is reasonable since we're likely to get one of each color when the numbers of yellow and blue balls are similar.